Proof. As mentioned before, through the self-similar change of variables 195 , 23 25 hold under the assumptions in Theorem 1. Take \(l=1+\theta+2m_1\) where \(m_1\) is any small positive constant. By 28 , for small \(\tilde{\varepsilon},\) \[\begin{align} \tilde{\varepsilon}^l=2^l\varepsilon^l e^{-\frac{A(1+\theta)}{1+X}}e^{-A\frac{2m_1}{1+X}}\leq 2^l\varepsilon^l e^{-A}e^{-Am_1}\leq \varepsilon^{10}e^{-A}, \end{align}\] where we have used \(\frac{2}{1+X}>1\) by 23 with \(\theta\) small and for any positive constants \(k\) and \(m_1\), \[\varepsilon^{-k}e^{-Am_1}\rightarrow 0\quad \text{ as}\quad \varepsilon\rightarrow0,\] since \(A\) is \(\varepsilon\) to some negative power. Then 26 implies \[\begin{align} \begin{aligned} &\|(\partial_zu_{bd},\partial_zv_{bd}) -(\partial_zu_B,\partial_zv_B)\|_{W_{x,y}^{2,\infty}(\{x=0\}\cup\{y=0\}\cup\{z=0\}\cap\partial \Omega)}\\ \leq& \varepsilon^9 e^{-A}[z\chi +e^{-\frac{3}{2} \mu z^2}(1-\chi)], \end{aligned} \end{align}\]by 28 and noting \(A\) is \(\varepsilon\) to some negative power. Note by 23 , \[\begin{align} e^{-A}\leq e^{-\frac{A}{x+1}}. \end{align}\] Then by 27 and 26 , for \(z\) large, \[\begin{align} |u_{bd}-u_B|\leq &\int_z^{+\infty} |\partial_z u_{bd}-\partial_z u_B|dz' \leq \varepsilon^9e^{-\frac{A}{x+1}} C\int_z^{+\infty}e^{-\frac{3}{2}\mu z^2}dz' \leq \varepsilon^9e^{-\frac{A}{x+1}} Ce^{-\frac{3}{2}\mu z^2}. \end{align}\] Moreover, by \(u=v=0\) at \(z=0\) and compatibility, it holds \[\begin{align} \partial_{x,y} u=\partial_{x,y} v= \partial_{x,y}\partial_{x,y} u=\partial_{x,y}\partial_{x,y} v=0 \quad at\quad z=0. \end{align}\] Hence, 27 and 26 imply \[\begin{align} \label{20250703-rd-main}\|(u_{bd},v_{bd}) -(u_B,v_B)\|_{W_{x,y}^{2,\infty}}\leq \varepsilon^9 e^{-\frac{A}{x+1}}C[z^2\chi +e^{-\frac{3}{2}\mu z^2}(1-\chi)], \end{align}\tag{29}\] Combining 26 , 29 , 1 and compatibility, we have \[\begin{align} \label{20250703-rd-main-1}\|(u_{bd},v_{bd}) -(u_B,v_B)\|_{W_{z}^{3,\infty}}\leq \varepsilon^9 Ce^{-\frac{A}{x+1}}. \end{align}\tag{30}\] On the other hand, by \(u=v=0\) at \(z=0\) and \(w=-\int_0^z\,\partial_xu+\partial_y v\,dz'\), the solution \(u\) to 1 with 27 13 satisfies \[\partial_z^2 u=\partial_z^3 u=0\quad \text{at}\quad z=0,\] by compatibility. Combining with 26 and 30 , it holds \[\begin{align} |\partial_zu_{bd} -\partial_zu_B|\leq \varepsilon^9e^{-\frac{A}{x+1}}C [z^2\chi +e^{-\frac{3}{2}\mu z^2}(1-\chi)]. \end{align}\] In addition, 24 and 26 imply \[\partial_zu_{bd},\,\,\partial_zv_{bd}\gtrsim e^{-\frac{3}{2}\mu z^2},\quad z>0.\] In a similar way, we can obtain other inequalities in ?? from 26 . ◻

Proof. If \(n=1\), then \(u_{n-1}=u_0=\bar{u}\) and the result holds. Now we consider \(n\geq 2.\)

By 44 , the growth rates of \(\bar{u}\) and its derivatives in Subsection 7.3, 227 and 35 , we have \[\begin{align} \label{250416commongr}\begin{aligned} &| \frac{\int_0^z \partial_{xy}^2u_{n-1}dz'}{u_{n-1}}|\leq C\bar{u}^{\frac{\alpha}{2}}, \quad|\partial_z( \frac{\int_0^z \partial_{y}v_{n-1}dz'}{v_{n-1}})|\leq C, \\& |\frac{\int_0^z \partial_{y}v_{n-1}dz'}{v_{n-1}}|\leq C\bar{u}\quad\text{in}\quad\Omega.\end{aligned} \end{align}\tag{48}\] and \[\begin{align} \label{250416-nabla1}\begin{aligned} | \nabla_{\xi}^{n-2}u_{n-1}-\nabla_{\tau_1}\bar{u}|=&|\nabla_{\xi}^{n-2}(u_{n-1}-\bar{u}) +(\nabla_{\xi}^{n-2}-\nabla_{\tau_1})\bar{u}|\leq \varepsilon^2C\phi_{1,1},\\ | \nabla_{\eta}^{n-2}u_{n-1}-\nabla_{\tau_1}\bar{u}|=& | \nabla_{\eta}^{n-2}u_{n-1}-\nabla_{\tau_2}\bar{u}|\leq \varepsilon^2C\phi_{1,1}. \end{aligned} \end{align}\tag{49}\] Here we recall the definition of \(\nabla_{\xi}^{n-2},\) \(\nabla_{\eta}^{n-2}\) in 31 and \[\nabla_{\xi}^{0}=\nabla_{\tau_1}, \nabla_{\eta}^{0}=\nabla_{\tau_2}.\]

Step 1 We will prove \[\begin{align} \label{250416-zz} |\partial_z^2(u_{n-1}-\bar{u})|\leq \varepsilon^{2}C\frac{1}{\bar{u}} \phi_{1,\alpha}\quad\text{in}\quad \Omega. \end{align}\tag{50}\]

By 22 , we have \[\begin{align} \begin{aligned} \partial_{z}^2u_{n-1}= &[u_{n-2}(\nabla_{\xi}^{n-2} u_{n-1} +(1+\tilde{q}_{n-2})\nabla_{\eta}^{n-2} u_{n-1} )-\partial_{z}u_{n-1} \partial_{z}(\frac{u_{n-1}}{u_{n-2}} )]\frac{u_{n-2}}{u_{n-1}},\\ \partial_{z}^2\bar{u}= &2\bar{u}\nabla_{\tau_1} \bar{u} \quad \text{in}\quad \Omega. \end{aligned} \end{align}\] Note by 44 , it holds \[\begin{align} \begin{aligned} |\partial_z (\frac{ u_{n-1}}{u_{n-2}})|=&|\partial_z (\frac{ u_{n-1}-u_{n-2}}{u_{n-2}})|\\ \leq &\frac{ |\partial_z u_{n-1}-\partial_zu_{n-2}|}{u_{n-2}}+\frac{\partial_zu_{n-2}|u_{n-2}-u_{n-1}|}{u_{n-2}^2} \\ \leq& C\varepsilon^{5}\frac{1}{u_{n-2}} \phi_{1,\alpha}\quad\text{in}\quad \Omega.\end{aligned} \end{align}\] Moreover, combining 49 , we complete the proof of 50 .

Step 2 We will prove \[\begin{align} \label{250416-znabla1} | \partial_z\nabla _{\eta,\xi} (u_{n-1}-\bar{u})|\leq \varepsilon^2 C\phi_{1,\alpha}\quad \text{in}\quad \Omega. \end{align}\tag{51}\] By 35 , we have \[\begin{align} \partial_z\nabla _{\eta} (u_{n-1}-\bar{u})= & \partial_z\partial_y(u_{n-1}-\bar{u})-\partial_z( \frac{\int_0^z \partial_{y}v_{n-1}dz'}{v_{n-1}})\partial_z(u_{n-1}-\bar{u}) \\&- \frac{\int_0^z \partial_{y}v_{n-1}dz'}{v_{n-1}}\partial_z^2(u_{n-1}-\bar{u}). \end{align}\] Then by 48 and 50 , we prove \[\begin{align} | \partial_z\nabla _{\eta} (u_{n-1}-\bar{u})|\leq \varepsilon^2 C\phi_{1,\alpha}\quad \text{in}\quad \Omega. \end{align}\] Similarly, we can prove \(| \partial_z\nabla _{\xi} (u_{n-1}-\bar{u})|\leq \varepsilon^2 C\phi_{1,\alpha}\) in \(\Omega.\) Then we complete the proof of 51 .

Step 3 We will prove \[\begin{align} \label{250416-nabla22} | \nabla _{\eta}\nabla _{\xi} (u_{n-1}-\bar{u})|\leq \varepsilon^2 C\phi_{1,\frac{\alpha}{2}}\quad \text{in}\quad \Omega. \end{align}\tag{52}\] By 35 , we have \[\begin{align} \begin{aligned} \nabla _{\eta} \nabla _{\xi}(u_{n-1}-\bar{u})=&\partial_{yx}^2(u_{n-1}-\bar{u})- \frac{\int_0^z \partial_{xy}^2u_{n-1}dz'}{u_{n-1}}\partial_z (u_{n-1}-\bar{u}) \\&+ \frac{\partial_yu_{n-1}\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}^2}\partial_z (u_{n-1}-\bar{u}) \\&- \frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\partial_{zy}^2 (u_{n-1}-\bar{u}) - \frac{\int_0^z \partial_{y}v_{n-1}dz'}{v_{n-1}}\partial_z\nabla _{\xi}(u_{n-1}-\bar{u}) \end{aligned} \end{align}\] Then by 44 , 48 and 51 , we complete the proof of 52 . The rest inequalities in this lemma can be proved similarly and then we skip the details. ◻

Proof. Set \[G=fe^{-Bx},\] where \[\begin{align} \label{largeB} B>\|c\|_{L^\infty((0,X]\times(0,Y]\times(0,z_0))}. \end{align}\tag{63}\]

By the assumption, \(G\leq 0\) on the boundary \(\big(\{x=0\}\cup\{y=0\}\cup\{z=0,z_0\}\big)\cap [0,X]\times[0,Y]\times[0,z_0].\) Hence, \(G\) does not attain a positive maximum on the boundary.

We now prove \(G\) does not attain any positive maximum in the interior \((0,X]\times(0,Y]\times(0,z_0)\) by contradiction. If \(G\) attains its positive maximum at some interior point \(p\in (0,X]\times(0,Y]\times(0,z_0),\) then \[\begin{align} \label{fp} f(p)>0, \end{align}\tag{64}\] and \(\partial_zG(p)=0, \partial_xG(p)\geq0, \partial_yG(p)\geq0\) and \(\partial_z^2G(p)\leq0.\) Hence \[\begin{align} \label{xzx}\begin{aligned} \nabla_\eta G(p)\geq&0,\quad \nabla_\xi G(p)\geq0,\quad \nabla_{\psi}G(p)=0, \\ \nabla_{\psi}^2G=&\frac{1}{u_{n-1}}\partial_z(\frac{1}{u_{n-1}}\partial_zG) =\frac{1}{u_{n-1}}( \frac{1}{u_{n-1}}\partial_z^2G- \frac{\partial_zu_{n-1}}{u_{n-1}^2}\partial_zG) \leq0\quad \text{at}\quad p,\end{aligned} \end{align}\tag{65}\] which implies that \[LG-cG\geq 0\quad \text{at} \quad p.\] However, by 63 and 64 , \[\begin{align} LG-cG=e^{-Bx}Lf+(-B-c)e^{-Bx}f<0\quad \text{at} \quad p, \end{align}\] which leads to a contradiction. Therefore, \(G\leq 0\) in \([0,X]\times[0,Y]\times[0,z_0]\), so is \(f.\) ◻

Proof. Set \[\begin{align} \label{hatpsi} \hat{\psi}_{z}=\min_{(x,y)\in[0,X]\times[0,Y]}\psi_{n-1}(x,y,z), \quad z\in[0,+\infty), \end{align}\tag{66}\] where \(\psi_{n-1}\) is defined in 37 . Then \[\begin{align} \label{hpsitoinf} \hat{\psi}_{z}\rightarrow+\infty, \quad z\rightarrow+\infty, \end{align}\tag{67}\] because \[\begin{align} \lim_{z\rightarrow+\infty}u_{n-1}(x,y,z)= 1, \quad (x,y)\in[0,X]\times[0,Y]. \end{align}\]

For any \(z_1\in(0,+\infty)\) and any \(z_2\in(z_1,+\infty),\) by the definition 66 , \(\hat{\psi}_{z_2}\) is a constant such that \[0<\hat{\psi}_{z_2}\leq\psi_{n-1}(x,y,z_2),\quad (x,y)\in[0,X]\times[0,Y].\] Set \[\begin{align} G(x,y,z)=&f(x,y,z)-\frac{M\psi_{n-1}^\alpha(x,y,z)}{\hat{\psi}_{z_2}^\alpha}e^{Bx} \quad \text{in} \quad [0,X]\times[0,Y]\times[0,z_2], \end{align}\] where \(\alpha\in(0,1)\) and \(B\) is a large positive constant such that \[B>\|c\|_{L^{\infty}( [0,X]\times[0,Y]\times[0,+\infty))}+\frac{1+\|b\|_{L^\infty((0,X]\times(0,Y]\times(0,\infty))}}{\delta}.\] Then \(G\leq 0\) on \(\{z=0,z_2\}\cup\{x=0\}\cup\{y=0\}\) and by ?? , 59 , 60 and 62 , \[\begin{align} L G(x,y,z)<&Lf(x,y,z)+( \frac{1+\|b\|_{L^\infty((0,X]\times(0,Y]\times(0,\infty))}}{\delta}-c-B)\frac{M\psi_{n-1}^\alpha(x,y,z)}{\hat{\psi}_{z_2}^\alpha}e^{Bx} \\ < &0 \quad \text{in} \quad (0,X]\times(0,Y]\times[0,z_2]. \end{align}\] By Lemma 3, we have \(G\leq 0\) in \([0,X]\times[0,Y]\times[0,z_2].\) Hence \(f(x,y,z_1)\leq \frac{M\psi_{n-1}^{\alpha}(x,y,z_1)}{\hat{\psi}_{z_2}^\alpha}e^{Bx}.\) Letting \(z_2\rightarrow+\infty,\) we have \(f(x,y,z_1)\leq0\) by 67 . Finally, since \(z_1\) is arbitrary, we have \(f\leq 0\) in \([0,X]\times[0,Y]\times[0,+\infty).\) ◻

Proof. We employ a proof by contradiction. If \(f-\phi\) attains its maximum at \(z_0,\) then \[\begin{align} & \partial_z^+(f-\phi)=\lim_{t\rightarrow0^+}\frac{(f-\phi)|_{z=z_0+t}-(f-\phi)|_{z=z_0}}{t}\leq 0,\\ &\partial_z^-(f-\phi)=\lim_{t\rightarrow0^+}\frac{(f-\phi)|_{z=z_0-t}-(f-\phi)|_{z=z_0}}{-t}\geq 0. \end{align}\] Hence, \(\partial_z^-(f-\phi)\geq\partial_z^+(f-\phi)\) at \(z_0\), since \(f\in C^{1}(\mathbf{R}_+)\). then \(\partial_z^-\phi\leq\partial_z^+\phi\) at \(z_0,\) which contradicts to ?? . The proof is completed. ◻

Proof. First, auxiliary function \(f-\varepsilon^{6.5}\phi_{2,1}\) cannot attain its maximum on the ridge \(\{\psi_{n-1}(x,y,z)=\delta\}\) by Proposition 4, since \(\partial_z^-\phi_{2,1}>\partial_z^+\phi_{2,1}\) at this ridge by noting \(\partial_z \psi_{n-1}=u_{n-1}\). Next, we claim \[\begin{align} L_{simple}(\varepsilon^{6.5}\phi_{2,1})> \phi_{2,0} \quad \text{in}\quad \Omega\setminus\{\psi_{n-1}(x,y,z)=\delta\}. \end{align}\] Then \(L_{simple}(f-\varepsilon^{6.5}\phi_{2,1})<0\) in \(\Omega\setminus\{\psi_{n-1}(x,y,z)=\delta\}.\) Hence, \(f-\varepsilon^{6.5}\phi_{2,1}\) cannot attain its maximum in \(\Omega\setminus\{\psi_{n-1}(x,y,z)=\delta\}\) by the same argument in the proof of the maximum principles in terms of vector fields.

We now prove the Claim as follows. See Figure 1. Take \[\begin{align} \Omega_s=&\{0\leq z\leq \varepsilon^m\}\cap\Omega\subset \{0\leq \psi_{n-1}(x,y,z)\leq \varepsilon^{2m}C\},\\ \Omega_b=&\{ z\geq \varepsilon^m\}\cap\Omega\subset \{ \psi_{n-1}(x,y,z)\geq \varepsilon^{2m}c\}, \end{align}\] where \(c\) and \(C\) are independent of \(\varepsilon.\) Then by the definition in 75 , we have \[L_{simple}\phi_{2,1}= \left\{ \begin{align}e^{-\frac{A}{x+1}}\alpha(1+\alpha)\psi_{n-1}^{\alpha-1}+ \frac{ A}{(1+x)^2}e^{-\frac{A}{x+1}}(\psi_{n-1}-\psi_{n-1}^{1+\alpha}),\quad & \psi_{n-1}(x,y,z)\leq \delta,\\ \frac{A}{(1+x)^2} e^{-\frac{A}{x+1}}\delta(1-\delta^{\alpha}),\quad & \psi_{n-1}(x,y,z)\geq \delta. \end{align}\right.\\\] Then \[\begin{align} L_{simple}(\varepsilon^{6.5}\phi_{2,1})> e^{-\frac{A}{x+1}} \quad \text{in}\quad \Omega_s \end{align}\] by taking \(m>\frac{6.5}{2(1-\alpha)}\) so that \(\varepsilon^{2m(1-\alpha)}<\varepsilon^{6.5}\) with \(\varepsilon\) being sufficiently small. Since for \(\delta\) small independent of \(\varepsilon,\) we have \[\psi_{n-1}-\psi_{n-1}^{1+\alpha}\geq \frac{1}{2}\psi_{n-1}\quad \text{for} \quad \psi_{n-1}(x,y,z)\leq \delta,\] for \(A\geq C\varepsilon^{-2m-6.5}\) with \(C\) being independent of \(\varepsilon,\) we have \[\begin{align} L_{simple}(\varepsilon^{6.5}\phi_{2,1})\geq \varepsilon^{6.5} \frac{A}{(x+1)^2}e^{-\frac{A}{x+1}}\frac{c}{2} \varepsilon^{2m}> \phi_{2,0} \quad \text{in} \quad\Omega_b\setminus\{\psi_{n-1}(x,y,z)=\delta\}. \end{align}\] ◻

Proof. Our goal is to prove \(g=u_{n}^2-\bar{u}^2-\varepsilon^{8} C{\epsilon_0}^3\phi_{2,0} -\varepsilon^7 \phi_{2,\beta}\leq 0\) in \(\Omega\cap\{0\leq y\leq Y^*\}\) by generalized maximum principle for functions with ridges.

Step 1 We will prove \(g\) can not attain positive maximum on \(\Omega\cap\{0\leq y\leq Y^*\}\setminus \{\text{ridges}\}.\)

We will prove by contradiction. By 72 and 81 , we have in \(\Omega\cap\{0\leq y\leq Y^*\}\), \[\begin{align} P_2 \big(u_{n}^2-\bar{u}^2-\varepsilon^{8} C{\epsilon_0}^3\phi_{2,0} \big) \leq R_0 , \end{align}\] where \(R_0\) is defined in 82 . By 228 , 229 and ?? in Lemma 18, we have in \(\Omega\cap\{0\leq y\leq Y^*\}\),\[\begin{align} |(\nabla_{\tau_1}+\nabla_{\tau_2} -\nabla_{\xi}-(1+\tilde{q})\nabla_{\eta} ) \bar{u}^2+(-\bar{u}\nabla_{n}^2+u_{n}\nabla_{\psi}^2 )\bar{u}^2|\leq C\bar{u}^2\phi_{1,0}+C\phi_{1,0}. \end{align}\] This implies \(|R_0|\leq C\phi_{1,0}\) where we note that by 199 , \[\begin{align} \label{lb0126g} c_0\epsilon_0\leq u_{n-1}\quad \text{in}\quad\Omega\cap\{0\leq y\leq Y^*\}. \end{align}\tag{84}\] Then by 72 and taking \(A\gg \varepsilon^{-7}\), it holds \[P_2 (u_{n}^2-\bar{u}^2-\varepsilon^{8} C{\epsilon_0}^3\phi_{2,0} -\varepsilon^7 \phi_{2,\beta} )<0\quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}\setminus \{\frac{\psi_{n-1}(x,y,z)}{\sqrt{x+1}}=\delta, N\},\] as explained in Example 1.

However, if \(g\) attains its positive maximum at some point \(p_0\in (0,X]\times(0,Y^*]\times(0,+\infty),\) then by the argument in the proof of maximum principle, we have \[P_2(g)|_{p_0}\geq 0,\] which leads to a contradiction. Then we complete step 1.

Step 2 We will prove \(g\) can not attain positive maximum on ridges.

\(g\) cannot attain its positive maximum on the ridges \(\{\frac{\psi_{n-1}(x,y,z)}{\sqrt{x+1}}=\delta, N\}\) by Proposition 4, since for any fixed \(x\), \[\partial_z^-(\varepsilon^{8} C{\epsilon_0}^3\phi_{2,0}+\varepsilon^7 \phi_{2,\beta})>\partial_z^+(\varepsilon^{8} C{\epsilon_0}^3\phi_{2,0}+\varepsilon^7 \phi_{2,\beta})\quad\text{at the ridges}.\]

Combining Step 1 and Step 2, \(g\) can not attain positive maximum on \(\Omega\cap\{0\leq y\leq Y^*\}\).

Next, by the boundary condition including the compatible condition 200 , i.e. \[\begin{align} |u_{n}^2-\bar{u}^2|\leq \varepsilon^{8} C{\epsilon_0}^3\phi_{2,0}\quad \text{at}\quad \{z=0\}\cap\overline{\Omega}, \end{align}\] we have \(g\leq 0\) on the boundary.

In summary, by the argument in the proof of Lemma 4[Maximum Principle in unbounded domain], we have \(g=u_{n}^2-\bar{u}^2-\varepsilon^{8} C{\epsilon_0}^3\phi_{2,0} -\varepsilon^7 \phi_{2,\beta}\leq 0\) in \(\Omega\cap\{0\leq y\leq Y^*\}\).

Similarly, we can prove \[[-(u_{n}^2-\bar{u}^2)-\varepsilon^{8} C{\epsilon_0}^3\phi_{2,0}] -\varepsilon^7 \phi_{2,\beta}\leq 0\quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}.\] ◻

Proof. By Lemma 5 and the prescribed boundary conditions on \(\{x=0\}\cup\{y=0\},\) \[\pm(u_{n}^2-\bar{u}^2)-\varepsilon^{6.5} \phi_{1,0}\leq 0\quad \text{on}\quad [0,X]\times[0,Y^*]\times[N,+\infty)\cap(\{x= 0\}\cup\{y=0\}\cup\{z= N\}),\] for small \(\varepsilon.\) Moreover, by taking \(A>>\varepsilon^{-6.5}\), 81 and 225 , we have \[\begin{align} P_2\big(\pm(u_{n}^2-\bar{u}^2)-\varepsilon^{6.5} \phi_{1,0}\big) =R_0 -\frac{1}{2}A\varepsilon^{6.5} \phi_{1,0}<0\quad \text{in}\quad (0,X]\times(0,Y^*]\times(N,+\infty) \end{align}\] where \(|R_0|\leq C\phi_{1,0}\) by the estimates in the proof of Lemma 5. Then by applying the maximum principle in the domain \([0,X]\times[0,Y^*]\times[N,+\infty),\) we have the desired result. ◻

Proof. For convenience, we take \(\delta_4\) small enough such that \(\psi_{n-1}(x,y,\delta_4)\leq \delta\) where \(\delta\) and \(\phi_{2,1}\) are defined in 75 .

First, we will prove by contradiction that \[\begin{align} \label{sm10126step1} u_{n}^2-\bar{u}^2 \leq \varepsilon^{8} C{\epsilon_0}^3\phi_{2,0}+ \varepsilon^{6.5} \phi_{2,1}\psi_{n-1}^\alpha \quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}\cap\{0\leq z\leq \delta_4\}. \end{align}\tag{85}\]

If 85 holds in \(\Omega\cap\{0\leq y\leq Y^*\}\cap\{0\leq z\leq \delta_4\},\) then it is proved.

Otherwise, there exists an interior point \(p_0\in(0,X]\times(0,Y^*]\times(0,\delta_4)\) such that \((u_{n}^2-\bar{u}^2-\varepsilon^{8} C{\epsilon_0}^3\phi_{2,0})\psi_{n-1}^{-\alpha}-\varepsilon^{6.5} \phi_{2,1}\) attains its positive maximum in \(\Omega\cap\{0\leq y\leq Y^*\}\cap\{0\leq z\leq \delta_4\}\) at \(p_0,\) since ?? holds on \(x=0,\) \(y=0\) and \(z=0,\delta_4,\) by the boundary data and Lemma 5. Then by applying the argument used for 65 , we have \[\begin{align} \label{p0sign} P_2[(u_{n}^2-\bar{u}^2-\varepsilon^{8} C{\epsilon_0}^3\phi_{2,0})\psi_{n-1}^{-\alpha}-\varepsilon^{6.5} \phi_{2,1}](p_0)\geq 0, \end{align}\tag{86}\] and \[\begin{align} \label{pp0} u_{n}^2-\bar{u}^2 - \varepsilon^{8} C{\epsilon_0}^3\phi_{2,0}>0\quad \text{at}\quad p_0. \end{align}\tag{87}\] On the other hand, for small \(\varepsilon,\) by ?? , 88 , 58 and 87 , we have \[\begin{align} P_2 \big((u_{n}^2-\bar{u}^2 -\varepsilon^{8} C{\epsilon_0}^3\phi_{2,0})\psi_{n-1}^{-\alpha}\big) \leq& |R_0|\psi_{n-1}^{-\alpha}-\alpha(\alpha+1)(u_{n}^2-\bar{u}^2- \varepsilon^{8} C{\epsilon_0}^3\phi_{2,0} ) u_n \psi_{n-1}^{-\alpha-2} \\&+2\alpha u_n \nabla_\psi (u_{n}^2-\bar{u}^2 -\varepsilon^{8} C{\epsilon_0}^3\phi_{2,0}) \psi_{n-1}^{-\alpha-1}\\&+\varepsilon^2C(u_{n}^2-\bar{u}^2- \varepsilon^{8} C{\epsilon_0}^3\phi_{2,0} )\psi_{n-1}^{-\alpha} \\ \leq& C\psi_{n-1}^{-\alpha-1}u_{n-1}^2\phi_{1,0}\quad \text{at}\quad p_0. \end{align}\] In fact, we have used the following: \[\begin{align} &|\nabla_\eta \psi_{n-1}^{-\alpha}|\leq \varepsilon^2C \psi_{n-1}^{-\alpha}, \quad\epsilon_0\leq Cu_{n-1}\quad \text{in } \quad \Omega\cap\{0\leq y\leq Y^*\}\cap\{0\leq z\leq \delta_4\} , \\&P_2 \big(u_{n}^2-\bar{u}^2-\varepsilon^{8} C{\epsilon_0}^3\phi_{2,0} \big) \leq R_0\quad \text{in } \quad \Omega\cap\{0\leq y\leq Y^*\}\cap\{0\leq z\leq \delta_4\} , \end{align}\] with \[\begin{align} \label{xz3} |R_0|\leq C\phi_{1,0} \end{align}\tag{88}\] from the proof of Lemma 5. Then by 72 and argument as explained in Example 1, taking \(A\gg \varepsilon^{-7}\) and \(\delta_4\) sufficiently small, for small \(\alpha\) with \(2\alpha<\frac{1}{2}\), it holds that \[P_2 \big((u_{n}^2-\bar{u}^2-\varepsilon^{8} C{\epsilon_0}^3\phi_{2,0} )\psi_{n-1}^{-\alpha} -\varepsilon^{6.5} \phi_{2,1} \big)(p_0)<0,\] which contradicts to 86 . Therefore, \((u_{n}^2-\bar{u}^2-\varepsilon^{8} C{\epsilon_0}^3\phi_{2,0})-\varepsilon^{6.5} \phi_{2,1}\psi_{n-1}^{\alpha}\leq 0\) in \(\Omega\cap\{0\leq y\leq Y^*\}\cap\{0\leq z\leq \delta_4\}\).

Similarly, we can prove \[\begin{align} -( u_{n}^2-\bar{u}^2 )\leq \varepsilon^{8} C{\epsilon_0}^3\phi_{2,0}+ \varepsilon^{6.5} \phi_{2,1}\psi_{n-1}^\alpha \quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}\cap\{0\leq z\leq \delta_4\}. \end{align}\] Hence, ?? holds in \(\Omega\cap\{0\leq y\leq Y^*\}\cap\{0\leq z\leq \delta_4\}.\)

Finally, 230 and 84 give ?? . ◻

Proof. By 83 , \[\begin{align} \begin{aligned} P_2(\nabla_{\psi}u_{n}^2 -\nabla_{n}\bar{u}^2) = R_1.\end{aligned} \end{align}\]

Step 1: We will prove \(| R_1|\leq C\phi_{1,0}.\)

Note that \[\frac{1}{2}\nabla_{\psi}u_{n}^2=\frac{u_{n}}{u_{n-1}}\partial_z u_{n}.\] By induction assumption and the definition of \(Y^*\), we have \[\begin{align} &| \nabla_{\eta}\tilde{q}|\leq \varepsilon^2 e^{-\frac{A}{x+1}}C,\quad |\nabla_{\psi}\tilde{q}| = |\frac{\partial_z\tilde{q}}{u_{n-1}}|\leq\frac{\varepsilon^5C}{u_{n-1}^2}\phi_{1,\alpha}, \\&|\nabla_{\eta}u_{n}^2|\leq Cu_{n-1}\min\{z+\epsilon_0,1\}e^{-\frac{(z+{\epsilon_0})^2}{x+1}\mu}. \end{align}\] Then by 228 in Appendix, \[\begin{align} |\nabla_{\eta}\tilde{q}\nabla_{\psi}u_{n}^2 -\nabla_{\psi}\tilde{q}\nabla_{\eta}u_{n}^2 +(\nabla_{\tau_1} + \nabla_{\tau_2}-\nabla_{\xi}-(1+\tilde{q})\nabla_{\eta} ) \nabla_{n}\bar{u}^2 | \leq C \phi_{1,0}. \end{align}\] Since \(\partial_z\nabla_{n}\bar{u}^2=2\partial_z^2\bar{u}\), by 205 , \(|\partial_z\nabla_{n}^2 \bar{u}^2|\leq Ce^{-\frac{(z+{\epsilon_0})^2}{x+1}\mu}\) and \(|\frac{2}{\bar{u}}\partial_z^2\bar{u}|=|\nabla_{n}^2 \bar{u}^2|\leq C\bar{u}e^{-\frac{(z+{\epsilon_0})^2}{x+1}\mu}.\) Hence, by 229 \[|(-\bar{u}\nabla_{n}^2+u_{n}\nabla_{\psi}^2 )\nabla_{n}\bar{u}^2 |\leq C\phi_{1,0}.\] Then by 78 , induction assumption and definition of \(Y^*,\) it holds that \[\begin{align} | \nabla_{\psi}^2u_n^2-\nabla_{n}^2\bar{u}^2|=| 2 \nabla_{\xi} u_{n}+2(1+\tilde{q})\nabla_{\eta}u_{n} - 4\nabla_{\tau_1}\bar{u}|\leq \varepsilon^2 C\phi_{1,1}. \end{align}\] Thus, we have \[\begin{align} |(\nabla_{n} \bar{u}^2)\frac{1}{2\bar{u}}(\nabla_{\psi}^2u_n^2-\nabla_{n}^2\bar{u}^2)| \leq |\frac{2\partial_z\bar{u}}{2\bar{u}}(\nabla_{\psi}^2u_n^2-\nabla_{n}^2\bar{u}^2)| \leq \varepsilon^2 C\phi_{1,0}. \end{align}\] By 78 , \(|\nabla_{\psi}^2u_{n}^2|\leq Cu_{n-1}e^{-\frac{(z+{\epsilon_0})^2}{x+1}\mu}.\) Then \[\begin{align} | \nabla_{\psi}^2u_{n}^2( \frac{1}{2u_{n}} \nabla_{\psi}u_{n}^2-\frac{1}{2\bar{u}}\nabla_{n}\bar{u}^2)|\leq Cu_{n-1} | \frac{1}{u_{n-1}}\partial_zu_{n}-\frac{1}{\bar{u}}\partial_z\bar{u}|\leq C\phi_{1,0}. \end{align}\] In summary, \(|R_1|\leq C\phi_{1,0},\) where \(R_1\) is defined in 83 .

Step 2: We will derive ?? .

Employing 225 and taking \(A\gg \varepsilon^{-6}\), by a similar argument on 83 as in the previous proof of Lemma 5 and Lemma 6, we have \[\begin{align} \frac{1}{2} | \nabla_{\psi}u_{n}^2-\nabla_{n}\bar{u}^2|=|\frac{u_n}{u_{n-1}}\partial_z u_n-\partial_z\bar{u}|\leq& \frac{d_0\varepsilon^6}{4} \phi_{2,1}+C\varepsilon^7\epsilon_0\phi_{2,0}\quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\},\\ \frac{1}{2} | \nabla_{\psi}u_{n}^2-\nabla_{n}\bar{u}^2|=|\frac{u_n}{u_{n-1}}\partial_z u_n-\partial_z\bar{u}|\leq& \frac{d_0\varepsilon^6}{4} \phi_{1,0}\quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}. \end{align}\] This completes the proof of the theorem. ◻

Proof. First, by 79 and 68 , \[\begin{align} \begin{aligned} P_1(\nabla_{\psi}u_{n}^2) =\bar{R} ,\end{aligned} \end{align}\] where \[|\bar{R}|=| -\nabla_{\eta}\tilde{q}\nabla_{\psi}u_{n}^2 +\nabla_{\psi}\tilde{q}\nabla_{\eta}u_{n}^2| \leq \varepsilon^2 C\phi_{1,0}e^{-\mu \frac{(z+\epsilon_0)^2}{1+x}}\quad\text{in}\quad(0,X]\times(0,Y^*]\times[0,\infty).\] Take \(N_1\) large depending on \(N\) such that \[\begin{align} \label{N10316-1} |\frac{2u_n}{u_{n-1}}-2 |<\frac{1}{10} \quad\text{in}\quad(0,X]\times(0,Y^*]\times(N_1,\infty), \end{align}\tag{89}\] and \[\begin{align} \label{N10316} - c_0e^{-\frac{3}{2}\mu (z+\epsilon_0)^2}+|\bar{R}|<0 \quad\text{in}\quad(0,X]\times(0,Y^*]\times(N_1,\infty), \end{align}\tag{90}\] where \(N\) is the constant in 43 . Then by 90 and Lemma 17, \[\begin{align} \label{0316intsign} P_1(5c_0e^{-\frac{3}{2}\mu(z+\epsilon_0)^2}-\nabla_{\psi}u_{n}^2 )<0 \quad\text{in}\quad(0,X]\times(0,Y^*]\times(N_1,\infty). \end{align}\tag{91}\]

Second, we analyze the boundary data. by ?? , for \(\varepsilon\) small depending on \(N_1,\) \[\begin{align} \label{0316psiz} \nabla_\psi u_n^2=\frac{2u_n}{u_{n-1}}\partial_z u_n \end{align}\tag{92}\] satisfies \[\begin{align} \label{mon0315bd} \nabla_\psi u_n^2\geq 5c_0e^{-\frac{3}{2}\mu (z+\epsilon_0)^2}\quad \text{on}\quad \Omega\cap\{0\leq y\leq Y^*\}\cap\{z= N_1\}, \end{align}\tag{93}\] and \[\begin{align} \label{mon0315bd-2-1} \partial_z u_n\geq 2c_0e^{-\frac{3}{2}\mu (z+\epsilon_0)^2}\quad \text{in}\quad \Omega\cap\{0\leq y\leq Y^*\}\cap\{z\leq N_1\}, \end{align}\tag{94}\] with \(c_0\) being small enough. Moreover, by 24 and ?? , \[\begin{align} \label{djdl0316ch} \nabla_\psi u_n^2\geq 5c_0e^{-\frac{3}{2}\mu (z+\epsilon_0)^2}\,\,\text{on} \,\,[0,X]\times[0,Y^*]\times[N_1,+\infty)\cap(\{x= 0\}\cup\{y=0\}). \end{align}\tag{95}\] Then by 93 and 95 , \(\nabla_\psi u_n^2\geq 5c_0e^{-\frac{3}{2}\mu (z+\epsilon_0)^2}\) on \([0,X]\times[0,Y^*]\times[N_1,+\infty)\cap(\{x= 0\}\cup\{y=0\}\cup\{z= N_1\}).\)

Finally, by the boundary condition, 91 and applying the maximum principle in \([0,X]\times[0,Y^*]\times[N_1,+\infty),\) we have \[\begin{align} \label{mon0315bd-1} 5c_0e^{-\frac{3}{2}\mu (z+\epsilon_0)^2}-\nabla_{\psi}u_{n}^2\leq 0 \quad \text{in}\quad \Omega\cap\{0\leq y\leq Y^*\}\cap\{z\geq N_1\}. \end{align}\tag{96}\] By 96 , 92 and 89 , we have \[\begin{align} \label{mon0315bd-2-1-final} \partial_z u_n\geq 2c_0e^{-\frac{3}{2}\mu (z+\epsilon_0)^2}\quad \text{in}\quad \Omega\cap\{0\leq y\leq Y^*\}\cap\{z\geq N_1\}. \end{align}\tag{97}\] Combining 94 and 97 , we have the conclusion. ◻

Proof. Step 1 We will prove ?? . By 101 , \(w=\nabla_{\eta,\xi}u_n\) satisfies \[\begin{align} \label{426}\begin{aligned} ( P_2+ P_3)(w-\nabla_{\tau_1}\bar{u}) =R_2 , \end{aligned} \end{align}\tag{103}\] where \(P_3\) contains 0th-order term and 1st-order derivative with respect to normal direction as follows, \[\begin{align} \begin{aligned} P_3 g:=- 2(\nabla_{\psi} u_n)\nabla_{\psi} g -\nabla_{\psi}^2 u_n g. \end{aligned} \end{align}\] By 227 , 77 for \(\nabla_{\psi}^2 u_n\) and \(\nabla_{\psi}^3 u_n\), 204 for \(\bar{u}\), 229 , Theorem 5 and Theorem 7, we have \[\begin{align} \label{r2sheng} | R_2|\leq \frac{C}{u_{n-1}^{2}}\phi_{1,0}. \end{align}\tag{104}\] Now we deal with \(P_3\) as follows. First, for \(\delta_1\in(0,\delta_0)\) small enough, \[\begin{align} -\nabla_{\psi}^2u_{n}=&-\frac{1}{u_{n-1}}\partial_z(\frac{1}{u_{n-1}}\partial_zu_n) =-\frac{1}{u_{n-1}^2}\partial_z^2u_n+\frac{1}{u_{n-1}^3}\partial_zu_{n-1}\partial_zu_n \\ \geq &\frac{c_0^2}{2u_{n-1}^3}\quad \text{in}\quad \Omega\cap\{0\leq y\leq Y^*\}\cap\{z\leq \delta_1\} , \end{align}\] where \(\delta_0\) and \(c_0\) are defined in 224 and 47 respectively which are independent of \(\alpha_0.\) Then for small positive constants \(\alpha_0, \delta_1\ll 1\), it holds \[\begin{align} \begin{aligned} &P_3\phi_{2,\alpha_0} +\frac{c_2\alpha_0(1-\alpha_0)}{8}\psi_{n-1}^{-\frac{3}{2 }}\phi_{2,\alpha_0}\\=&-2\nabla_{\psi}u_{n}\nabla_{\psi}\phi_{2,\alpha_0} -(\nabla_{\psi}^2u_{n})\phi_{2,\alpha_0}+\frac{c_2\alpha_0(1-\alpha_0)}{8}\psi_{n-1}^{-\frac{3}{2 }}\phi_{2,\alpha_0}\\=&( -2\alpha_0\frac{\partial_zu_n}{u_{n-1}\psi_{n-1}} -\frac{1}{u_{n-1}^2}\partial_z^2u_n +\frac{1}{u_{n-1}^3}\partial_zu_{n-1}\partial_zu_n +\alpha_0\frac{c_2(1-\alpha_0)}{8}\psi_{n-1}^{-\frac{3}{2 }})\phi_{2,\alpha_0} \\ \geq & 0 \quad \text{in}\quad \Omega\cap\{0\leq y\leq Y^*\}\cap\{z\leq \delta_1\} , \end{aligned} \end{align}\] where we have used the fact that for \(z\in[0, \delta_1]\), if \(\frac{2\partial_z u_n}{u_{n-1}\psi_{n-1}}\geq \frac{c_2(1-\alpha_0)}{8}\psi_{n-1}^{-\frac{3}{2 }},\) then for \(\alpha_0\ll 1,\) it holds \[\begin{align} |\alpha_0\frac{\partial_zu_n}{u_{n-1}\psi_{n-1}}|\leq \alpha_0\frac{C_0}{u_{n-1}\psi_{n-1}}\leq \alpha_0 (\frac{16C_0}{c_2(1-\alpha_0)})^2\frac{C_0}{u_{n-1}^3}\leq \frac{c_0^2}{2u_{n-1}^3}. \end{align}\] Note that \(c_2\) is defined in 224 and \(c_0, C_0\) are defined in 47 . Moreover, for some positive constants \(\lambda\) and \(c,\) \[\begin{align} \phi_{2,\alpha_0} \psi_{n-1}^{-\frac{3}{2}}\geq \lambda e^{-\frac{A}{1+x}} \psi_{n-1}^{\alpha_0-\frac{3}{2}}\geq ce^{-\frac{A}{1+x}} u_{n-1}^ {2\alpha_0-3}\quad \text{in}\quad \Omega\cap\{0\leq y\leq Y^*\}\cap\{z\leq \delta_1\} \end{align}\] and \(2\alpha_0-3<-2\) for \(\alpha_0<\frac{1}{2}.\)

On the other hand, in \(\Omega\cap\{0\leq y\leq Y^*\}\cap\{z\geq\delta_1\}\), by 58 , we have \[\begin{align} |\nabla_{\psi}u_{n}\nabla_{\psi}\phi_{2,\alpha_0}| =&|\frac{\partial_zu_n}{u_{n-1}}\nabla_{\psi}\phi_{2,\alpha_0}| =|-\frac{9}{5}\frac{\psi_{n-1}}{x+1}\mu\frac{\partial_zu_n}{u_{n-1}}\phi_{2,\alpha_0}|\\ \leq &C|\psi_{n-1}\partial_zu_n|\phi_{2,\alpha_0} \leq C\phi_{2,\alpha_0} \quad\text{in}\quad \Omega\cap\{0\leq y\leq Y^*\}\cap\{\frac{\psi_{n-1}}{\sqrt{x+1}}\geq N\} , \end{align}\]and \[\begin{align} |\nabla_{\psi}u_{n}\nabla_{\psi}\phi_{2,\alpha_0}| =&|\frac{\partial_zu_n}{u_{n-1}}\nabla_{\psi}\phi_{2,\alpha_0}| \leq C\phi_{2,\alpha_0} \,\,\text{in}\,\, \Omega\cap\{0\leq y\leq Y^*\} \cap\{z\geq\delta_1\}\cap\{\frac{\psi_{n-1}(x,y,z)}{\sqrt{x+1}}\leq N\} , \end{align}\] where we have used the facts that \(\nabla_\psi x=\frac{1}{u_{n-1}}\partial_z x=0\) and for some positive constants \(c\) and \(C,\) \[u_{n-1}\geq c_0\delta_1,\quad |\partial_zu_n|\leq Ce^{-c\frac{\psi_{n-1}^2(x,y,z )}{x+1}\mu}\quad \text{in}\quad \Omega\cap\{0\leq y\leq Y^*\}\cap\{z\geq\delta_1\} ,\] by 56 , 57 and the relation \(c\leq \frac{\psi_{n-1}}{z+\epsilon_0}\leq C\) in \(\Omega\cap\{0\leq y\leq Y^*\}\cap\{z\geq\delta_1\} .\) Then \(|P_3\phi_{2,\alpha_0}|\leq C\phi_{2,\alpha_0}\) in \(\Omega\cap\{0\leq y\leq Y^*\}\cap\{z\geq\delta_1\} .\)

Hence, by taking \(A\gg \frac{1}{\varepsilon^5}\) and by 103 and 224 , \[\begin{align} ( P_2+ P_3)(\pm(\nabla_{\eta,\xi}u_n-\nabla_{\tau_1}\bar{u})-C\varepsilon^6\epsilon_0^2\phi_{2,0} -\varepsilon^5 \phi_{2,\alpha_0})\leq 0, \end{align}\] in \(\Omega\cap\{0\leq y\leq Y^*\}\setminus\{\text{ridges of barrier functions}\}.\) Then by a similar argument as in the previous proof, we obtain ?? .

Step 2 We will prove ?? . By ?? , we only need to prove it for \(z\) large.

By ?? and the prescribed boundary conditions on \(\{x=0\}\cup\{y=0\},\) \[\pm(\nabla_{\eta,\xi}u_n-\nabla_{\tau_1} \bar{u})-C\varepsilon^{5} \phi_{1,0}\leq 0\quad \text{on}\quad [0,X]\times[0,Y^*]\times[N,+\infty)\cap(\{x= 0\}\cup\{y=0\}\cup\{z= N\}),\] for small \(\varepsilon.\) Next, by 103 , we have \[\begin{align} \begin{aligned} P_2(\nabla_{\eta,\xi}u_n-\nabla_{\tau_1}\bar{u}) =\hat{R}_2 \quad\text{in}\quad(0,X]\times(0,Y^*]\times(N,+\infty), \end{aligned} \end{align}\] where we note \(|\hat{R}_2|\leq C\phi_{1,0}\) in \((0,X]\times(0,Y^*]\times(N,+\infty).\) Hence, by 225 and taking \(A>>\varepsilon^{-5},\) we have \[\begin{align} \begin{aligned} P_2(\nabla_{\eta,\xi}u_n-\nabla_{\tau_1}\bar{u}-C\varepsilon^{5} \phi_{1,0}) <0 \quad\text{in}\quad(0,X]\times(0,Y^*]\times(N,+\infty). \end{aligned} \end{align}\] Then applying the maximum principle in the domain \([0,X]\times[0,Y^*]\times[N,+\infty),\) we obtain ?? . ◻

Proof of Theorem 12. Step 1 Equation of \(\nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\tau_1}^2\bar{u}\)

By 105 and 107 , we have \[\begin{align} \label{1104-1}\begin{aligned} 0=&P_1( \nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\tau_1}^2\bar{u})-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_z( \nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\tau_1}^2\bar{u})+\tilde{c}( \nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\tau_1}^2\bar{u})\\&+(1+\tilde{q})\partial_{y}(K\partial_zu_n) +f_{12}^u,\end{aligned} \end{align}\tag{113}\] where by 39 , \[\begin{align} \label{f12u}\begin{aligned} f_{12}^u=&f_{12}^{u-\bar{u}}-(1+\tilde{q})\frac{\int_0^z \partial_{y}v_{n-1}dz'}{v_{n-1}}\partial_{z}(K\partial_zu_n)\\ =&f_{12}^{u-\bar{u}}-(1+\tilde{q})\frac{\int_0^z \partial_{y}v_{n-1}dz'}{v_{n-1}}K\partial_z^2u_n-(1+\tilde{q})\frac{\int_0^z \partial_{y}v_{n-1}dz'}{v_{n-1}}\partial_{z}K\partial_zu_n \\ =&f_{12}^{u-\bar{u}}-(1+\tilde{q})\frac{\int_0^z \partial_{y}v_{n-1}dz'}{v_{n-1}}K\partial_z^2u_n \\&-(1+\tilde{q})\frac{\int_0^z \partial_{y}v_{n-1}dz'}{v_{n-1}}\partial_zu_n(-\nabla_{\xi}(\frac{\nabla_{\eta}\tilde{q}}{1+\tilde{q}}) -K\frac{\partial_z u_{n-1}}{u_{n-1}}).\end{aligned} \end{align}\tag{114}\] Recall \(f_{12}^{u-\bar{u}}\) in 108 and \[\begin{align} \label{utan2c}\begin{aligned} \tilde{c}=&(\nabla_{\eta}\tilde{q}-\frac{ \nabla_{\xi}\tilde{q}}{1+\tilde{q}} )-\frac{1}{u_{n-1}}\partial_z(\frac{\partial_zu_{n} }{u_{n-1}})\\ =&(\nabla_{\eta}\tilde{q}-\frac{ \nabla_{\xi}\tilde{q}}{1+\tilde{q}} )-\frac{\partial_z^2u_{n} }{u_{n-1}^2}+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3} \\:=&c_1+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}.\end{aligned} \end{align}\tag{115}\] Then by 41 , we have \[\begin{align} \label{cf12u} |c_1+\frac{\partial_z^2u_{n} }{u_{n-1}^2}|\leq \frac{C \phi_{1,0}}{u_{n-1}^2}, \quad |f_{12}^{u}|\leq C\varepsilon^2\phi_{1,0}+\frac{\varepsilon^2C\phi_{1,0}}{u_{n-1}^{3-\alpha}}. \end{align}\tag{116}\]

Step 2 Auxiliary function \(f\) and the goal

For any small positive constant \[\begin{align} \label{02072025} \gamma\ll \varepsilon^3e^{-A}\epsilon_0\leq \varepsilon^3e^{-\frac{A}{x+1}}\epsilon_0, \end{align}\tag{117}\] set \[\begin{align} \label{fplus} f= ( \nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\tau_1}^2\bar{u})-\frac{\varepsilon^2}{2}d_0^2\phi_{1,\frac{\alpha}{2}}-\gamma \quad \text{in} \quad[0,X]\times[0,Y]\times[0,+\infty). \end{align}\tag{118}\] Goal: We will prove \[\begin{align} \label{0323goalfneg} f\leq 0 \quad \text{in} \quad [0,X]\times[0,Y^*]\times[0,+\infty). \end{align}\tag{119}\] Then letting \(\gamma\) go to \(0\), we have \[\begin{align} \label{0324fremovegam} ( \nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\tau_1}^2\bar{u})-\frac{\varepsilon^2}{2}d_0^2\phi_{1,\frac{\alpha}{2}} \leq 0 \quad \text{in} \quad[0,X]\times[0,Y^*]\times[0,+\infty). \end{align}\tag{120}\]

We will employ a proof by contradiction. i.e. If 119 fails, then there exists a maximum “time" \[\begin{align} \label{ystar0220} y_*:=\max \{y_0\in[0,Y^*]|f(x,y,z)\leq 0,\quad \text{in} \quad [0,X]\times[0,y_0]\times[0,+\infty)\}, \end{align}\tag{121}\] such that \[\begin{align} \label{0323-ymaxt} y_*\in(0,Y^*), \end{align}\tag{122}\] by the boundary condition. In the following, we will prove that this leads to a contradiction.

As mentioned in the introduction, \(\partial_y (K\partial_z u_n)\) in 113 contains loss of tangential derivative which is one order higher than the induction assumptions. Hence, instead of bounding \(\partial_y K\), our analysis takes advantage of the sign of \(\partial_y f\) as follows.

Step 3 The sign of \(f\) and \(\partial_y f\) on specific domains

Firstly, by the boundary condition, we have \(f\leq -\gamma\) on \(z=0\). By continuity, there exists a positive constant \(z_0\) such that \[\begin{align} \label{tldM0126} f\leq-\frac{\gamma}{2} \quad \text{in} \quad [0,X]\times[0,Y^*]\times[0,z_0]. \end{align}\tag{123}\] Moreover, since \(\nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\tau_1}^2\bar{u}\rightarrow 0\) as \(z\) tends to infinity, there exists a large positive constant \(N_1\) such that \[\begin{align} \label{Ngamm} f+\frac{\gamma}{2}<0\quad \text{in} \quad[0,X]\times[0,Y]\times[N_1,+\infty). \end{align}\tag{124}\]

In addition, by the boundary condition, \[|\nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\tau_1}^2\bar{u}|-\frac{\varepsilon^2}{8}d_0^2(\phi_{1,\frac{\alpha}{2}})^2\leq 0\quad \text{on}\quad \{x=0\}\times[0,Y]\times[0,N_1].\]

Note \[\begin{align} \phi_{1,\frac{\alpha}{2}} (x,z) \geq \min_{[0,X]\times[0,N_1]}\phi_{1,\frac{\alpha}{2}} (x,z)>0, \quad (x,z)\in [0,X]\times[0,N_1]. \end{align}\] Then for the positive constant \(l=\min_{[0,X]\times[0,N_1]}\frac{\varepsilon^2}{16} d_0^2 \phi_{1,\frac{\alpha}{2}}>0,\) there exists a constant \(x_0\in(0,X)\) such that \(|\nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\tau_1}^2\bar{u}|-\frac{\varepsilon^2}{8}d_0^2(\phi_{1,\frac{\alpha}{2}})^2<l\) in \([0,x_0]\times[0,Y]\times[0,N_1].\) Hence, \(|\nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\tau_1}^2\bar{u}|-\frac{\varepsilon^2}{2}d_0^2\phi_{1,\frac{\alpha}{2}}< l+\frac{\varepsilon^2}{8}d_0^2(\phi_{1,\frac{\alpha}{2}})^2-\frac{\varepsilon^2}{2}d_0^2\phi_{1,\frac{\alpha}{2}}\leq-\frac{\varepsilon^2}{4}d_0^2 \phi_{1,\frac{\alpha}{2}}\). Then \[\begin{align} \label{x0} f<-\frac{\varepsilon^2}{4}d_0^2 \phi_{1,\frac{\alpha}{2}}<0\quad \text{in} \quad [0,x_0]\times[0,Y]\times[0,N_1]. \end{align}\tag{125}\]

In this step, we will study the properties of the function \(f\). First of all, we consider the signs of \(f\) and \(\partial_y f\) on the plane \(\{y=y_*\}.\)

Set \[D_{y_*}=[x_0,X]\times\{y=y_*\}\times[z_0,N_1],\] where \(z_0\) is defined in 123 . Then by the definition of \(y_*\) in 121 , we have \[f(p)\leq0,\quad p\in D_{y_*}.\]

For every \(p\in D_{y_*}\) , there are two possibilities.

(i) If \(f(p)<0,\) then there exists a positive constant \(r_p<\min\{z_0,x_0\}\) such that \[f\leq \frac{f(p)}{2} <0 \quad \text{in}\quad B_{r_p}(p)\cap D_{y_*},\] where \(B_{r_p}(p)\subseteq \mathbf{R}^2_{x,z}\times\{y=y_*\}\) and \(\mathbf{R}^2_{x,z}=\{(x,z)\in\mathbf{R}^2\}\).

(ii) If \(f(p)=0,\) then \(\partial_y f(p)\geq 0\) by the definition of \(y_*\) in 121 . Then there exists a positive constant \(r_p<\min\{z_0,x_0\}\) such that \[\partial_y f\geq -\frac{1}{100} \gamma \min_{[0,X]\times[0,N_1]} \phi_{1,\frac{\alpha}{2}} \quad \text{in}\quad B_{r_p}(p)\cap D_{y_*},\] where \(B_{r_p}(p)\subseteq \mathbf{R}^2_{x,z}\times\{y=y_*\}\).

Since \(\{B_{r_p}(p)\}_{p\in D_{y_*}}\) is an open covering of the compact set \(D_{y_*},\) there exists a finite subcover of \(D_{y_*}\) such that \[D_{y_*}\subseteq\big(\cup_{i\in\{1,\cdots,n_0\}}B_{r_{p_i^-}}(p_i^-)\big)\bigcup\big(\cup_{i\in\{1,\cdots,n_1\}}B_{r_{p_i^+}}(p_i^+)\big),\] where \(p_i^-\) denotes the point where \(f(p_i^-)<0\) and \(p_i^+\) denotes the point where \(\partial_y f(p_i^+)\geq 0\).

According to the choice of \(r_{p},\) we have, for some \(i_0\in\{1,\cdots,n_0\},\) \[\begin{align} f\leq \max_{i\in\{1,\cdots,n_0\}}\{\frac{f(p_i^-)}{2} \}=\frac{f(p_{i_0}^-)}{2}<0, \quad & D_{y_*}\cap\big( \cup_{i\in\{1,\cdots,n_0\}}B_{r_{p_i^-}}(p_i^-)\big),\\ \partial_y f\geq -\frac{1}{100} \gamma \min_{[0,X]\times[0,N_1]} \phi_{1,\frac{\alpha}{2}}, \quad & D_{y_*}\cap\big(\cup_{i\in\{1,\cdots,n_1\}}B_{r_{p_i^+}}(p_i^+)\big). \end{align}\] Then take a constant \(y_2\) with \(y_2-y_*\) small such that \(0<y_2-y_*\ll \min\{1,Y^*-y_*\}.\) By the continuity, \[\begin{align} \label{sginf}\begin{aligned} f+\gamma \phi_{1,\frac{\alpha}{2}}(y-y_*)\leq& \max\{ \frac{f(p_{i_0}^-)}{4},\,\,-\min_{[0,x_0]\times[0,N_1]}\frac{\varepsilon^2d_0^2}{4} \phi_{1,\frac{\alpha}{2}},\,\,-\frac{\gamma}{2}\}+\gamma \phi_{1,\frac{\alpha}{2}}(y_2-y_*) \\ <&0 \quad \quad\quad\quad\quad \text{in} \quad D_1 \times[y_*,y_2],\\ \partial_y (f+\gamma \phi_{1,\frac{\alpha}{2}}(y-y_*))\geq&0 \quad\quad\quad\quad\quad \text{in} \quad D_2\times[y_*,y_2],\end{aligned} \end{align}\tag{126}\] where the negative upper bound of \(f\) in \([0,x_0]\) is given in 125 , the negative upper bound of \(f\) in \([0,z_0]\) is given in 123 . Here, \[\begin{align} D_1=&\big(D_{y_*}\cap\cup_{i\in\{1,\cdots,n_0\}}B_{r_{p_i^-}}(p_i^-)\big)\bigcup [0,x_0]\times\{y=y_*\}\times[0,N_1]\bigcup [0,X]\times\{y=y_*\}\times[0,z_0], \\D_2=&\big(D_{y_*}\cap\cup_{i\in\{1,\cdots,n_1\}}B_{r_{p_i^+}}(p_i^+)\big). \end{align}\]

Note that \[\begin{align} \label{d1d2} D_1\cup D_2=[0,X]\times[0,N_1]\subseteq \mathbf{R}^2_{x,z}, \end{align}\tag{127}\] and \(D_i \times[y_*,y_2],\) \(i=1,2\) are the cylinders with bottom on the plane \(\{y=y_*\}\) and height being \(y_2-y_*.\)

Step 4 Equation of \(f\)

We now turn to estimate \[P_1 f-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_zf+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}f+c_1f.\] By the definition of \(f\) in 118 and straightforward calculation, we have \[\begin{align} \label{520}\begin{aligned} &P_1 f-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_zf+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}f+c_1f \\ =& -(1+\tilde{q})\partial_{y}(K\partial_zu_n) -f_{12}^u-\gamma(\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}+c_1) \\&-[P_1 \phi_{1,\frac{\alpha}{2}}-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_z\phi_{1,\frac{\alpha}{2}}+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}\phi_{1,\frac{\alpha}{2}}+c_1\phi_{1,\frac{\alpha}{2}}]\frac{\varepsilon^2}{2}d_0^2 \\ =& -(1+\tilde{q})\partial_{y}(K\partial_zu_n) -f_{12}^u-\gamma(\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}+c_1) \\&-[P_1 \phi_{1,\frac{\alpha}{2}}-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_z\phi_{1,\frac{\alpha}{2}}+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}\phi_{1,\frac{\alpha}{2}}+c_1\phi_{1,\frac{\alpha}{2}}](\frac{\varepsilon^2}{4}+\frac{\varepsilon^2}{4})d_0^2 ,\end{aligned} \end{align}\tag{128}\] where \(c_1\) is given in 115 . Taking \(A\) sufficiently large and the positive constant \(\alpha\) sufficiently small such that \[\begin{align} 0< \partial_z\phi_{1,\frac{\alpha}{2}}=\frac{\alpha}{2}\phi_{1,\frac{\alpha}{2}}\frac{1}{z+\epsilon_0} \leq \frac{ c_0 }{2u_{n-1}}\phi_{1,\frac{\alpha}{2}}\leq \frac{ \partial_z u_{n-1}}{2u_{n-1}}\phi_{1,\frac{\alpha}{2}}, \quad z\in[0,\delta-\epsilon_0], \end{align}\] where \(\delta\) and \(c_0\) are defined in 45 and 47 respectively, it holds \[\begin{align} \label{52302072025}\begin{aligned} &\frac{\varepsilon^2}{4}d_0^2[P_1 \phi_{1,\frac{\alpha}{2}}-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_z\phi_{1,\frac{\alpha}{2}}+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}\phi_{1,\frac{\alpha}{2}}+c_1\phi_{1,\frac{\alpha}{2}} ] \\ \geq&\varepsilon^2d_0^2(\frac{3\lambda}{u_{n-1}^3}+\frac{A}{5}c_2)\phi_{1,\frac{\alpha}{2}} >0,\quad z\in[0,+\infty),\end{aligned} \end{align}\tag{129}\] where \(\lambda\) is a constant with \(0<3\lambda\leq \frac{c_0^2}{8}\), \(c_2\) is defined in 224 . And we note that \(\partial_z^-\phi_{1,\frac{\alpha}{2}}>\partial_z^+\phi_{1,\frac{\alpha}{2}}\) at the ridges. By 116 and by taking \(\delta_2\) small enough such that \(\frac{\delta_2^{\frac{\alpha}{2}}}{\varepsilon }\ll d_0^2,\) we have, \[\begin{align} |f_{12}^{u}|\leq \frac{\varepsilon^3Cd_0^2\phi_{1,0}}{u_{n-1}^{3-\frac{\alpha}{2}},}\quad z\in[0,\delta_2]. \end{align}\]

Then for \(\gamma\ll \varepsilon^3e^{-A}\epsilon_0\leq\varepsilon^3 e^{-\frac{A}{x+1}}Cu_{n-1},\) \(\varepsilon\ll d_0^2\ll1\) and \(A\gg \frac{1}{\delta_2^{3}}\), we have \[\begin{align} \label{74614}\begin{aligned} &\partial_{x}f+(1+\tilde{q})\partial_{y}f -\frac{1}{u_{n-1}}\partial_{z} (\frac{u_n}{u_{n-1}}\partial_{z}f) \\&+(-\tilde{b}-\frac{\partial_zu_{n} }{u_{n-1}^2})\partial_zf+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}f+c_1f \\=&P_1 f-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_zf+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}f+c_1f \\ \leq &-\varepsilon^2d_0^2(\frac{2\lambda}{u_{n-1}^3}+\frac{A}{6}c_2)\phi_{1,\frac{\alpha}{2}} -(1+\tilde{q})\partial_{y}(K\partial_zu_n+(2-k_1)\varepsilon^{2} e^{-\frac{A}{x+1}}C_1) \\&-[P_1 \phi_{1,\frac{\alpha}{2}}-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_z\phi_{1,\frac{\alpha}{2}}+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}\phi_{1,\frac{\alpha}{2}}+c_1\phi_{1,\frac{\alpha}{2}}]\frac{\varepsilon^2}{4}d_0^2 \end{aligned} \end{align}\tag{130}\] where the function \(\tilde{b}\) is given in 70 and the two constants \(C_1\) and \(k_1\) are chosen as follows. For \(\varepsilon\) small enough, we have \[\begin{align} \label{k0} 2 (1+\tilde{q})\geq 2-k_0 \end{align}\tag{131}\] for some small positive constant \(k_0\ll 1\). \(C_1\) independent of \(\varepsilon\) and \(k_0< k_1\ll 1\) are two positive constants chosen such that \(C_1\gg C_2\) where the constant \(C_2\) is defined in 41 and \[\begin{align} \label{Kplus} (2-k_2)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1< (1+\tilde{q})( K\partial_zu_n+(2-k_1)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1)\leq\varepsilon^{2}e^{-\frac{A}{x+1}}C_1 (2-k_0), \end{align}\tag{132}\] with \(0<k_0< k_1<k_2\ll 1\).
Step 5 Auxiliary function \(F\) and proof of 21

Set \[\begin{align} \label{fmod}\begin{aligned} F=& \nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\tau_1}^2 \bar{u}-\frac{\varepsilon^2}{2}d_0^2\phi_{1,\frac{\alpha}{2}}-\gamma -2\varepsilon^{2}e^{-\frac{A}{x+1}}C_1+2 \varepsilon^{2}e^{-\frac{A}{x+1}}C_1\zeta(y) \\&+\gamma \phi_{1,\frac{\alpha}{2}}(y-y_*), \end{aligned} \end{align}\tag{133}\] in \([0,X]\times[y_*, y_2]\times[0,+\infty),\) where \[\label{zeta}\zeta(y)= \left\{ \begin{align} 1,\quad & y_*\leq y\leq y_1,\\ 1-\frac{y-y_1}{y_2-y_1},\quad & y_1< y\leq y_2. \end{align}\right.\\\tag{134}\] Here \(y_2<Y^*\) is defined in 126 and \(y_1\) will be determined as follows.

Take \(y_1\in(y_*,y_2)\) to be a positive constant so that \(y_2-y_1\) is sufficiently small to satisfy \[\begin{align} \label{y2-y1} 2 \varepsilon^{2}e^{-A}C_1\frac{1}{y_2-y_1}\gg\max_{[0,X]\times[0,Y^*]\times[0,N_1]} |\partial_y f|+\gamma |\phi_{1,\frac{\alpha}{2}}|. \end{align}\tag{135}\] This implies \[\begin{align} \label{k3dyF} 1-k_3\leq \frac{-\partial_y F}{d(x)}\leq 1+k_3\quad \text{in} \quad (y_1,y_2], \end{align}\tag{136}\] by \(e^{-\frac{A}{x+1}}\geq e^{-A},\) where \[\begin{align} \label{ddyzeta} d(x):=\frac{2 \varepsilon^{2}e^{-\frac{A}{x+1}}C_1}{y_2-y_1}. \end{align}\tag{137}\] \(k_3\ll 1\) is chosen small enough such that \[\begin{align} \label{kcomb} \frac{(1+k_3)(2-k_0)}{2}\leq (1-k_3)(2-k_2). \end{align}\tag{138}\]

Moreover, since \(|\phi_{1,\frac{\alpha}{2}}|\) is bounded, we have for small \(y_2-y_*\), \[-2\varepsilon^{2} e^{-\frac{A}{x+1}}C_1+2 \varepsilon^{2} e^{-\frac{A}{x+1}}C_1\zeta(y) +\gamma \phi_{1,\frac{\alpha}{2}}(y-y_*)\leq \frac{\gamma}{2},\quad y\in[y_*,y_2].\] Thus, by 124 we have \[\begin{align} \label{capfn1} F<0 \quad \text{ in} \quad [0,X]\times[y_*,y_2]\times[N_1,+\infty). \end{align}\tag{139}\] In addition, by the definition of \(y_*\) in 121 , there exists a point \((x^{1},y^{1},z^{1})\in \Omega\) such that \(f(x^{1},y^{1},z^{1})>0\) with \(y_*<y^{1}\leq y_1.\)

In this step, our goal is to prove \(F\leq 0\) in \([y_*,y_2]\) which implies \[\nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\tau_1}^2 \bar{u}-\frac{\varepsilon^2}{2}d_0^2\phi_{1,\frac{\alpha}{2}}-\gamma +\gamma \phi_{1,\frac{\alpha}{2}}(y-y_*)\leq 0 \quad \text{in} \quad [0,X]\times[y_*,y_1]\times[0,N_1],\] because \[\begin{align} F=f+\gamma \phi_{1,\frac{\alpha}{2}}(y-y_*) \quad \text{in} \quad [0,X]\times[y_*,y_1]\times[0,N_1]. \end{align}\] In particular, we have \(\nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\tau_1}^2 \bar{u}-\frac{\varepsilon^2}{2}d_0^2\phi_{1,\frac{\alpha}{2}}-\gamma \leq 0\) in \([y_*,y_1]\) which leads to a contradiction to the definition of \(y_*.\)

Since \(-2\varepsilon^2 e^{-\frac{A}{x+1}} C_1+2\varepsilon^2 e^{-\frac{A}{x+1}}C_1\zeta \leq 0,\) we have by 126 and 135 that \[\begin{align} \begin{aligned} F<0 \quad & \text{in} \quad D_1 \times[y_*,y_2],\\ \partial_y F\geq0 \quad & \text{in} \quad D_2\times[y_*,y_1),\\ \partial_y F<0\quad & \text{in} \quad [0,X]\times(y_1,y_2]\times[0,N_1].\end{aligned} \end{align}\] Hence, by 127 , \[\begin{align} \label{capF}\begin{aligned} \partial_y F_+\geq &0\quad \text{in} \quad [0,X]\times[y_*,y_1)\times[0,N_1],\\ \partial_y F_+\leq &0\quad \text{in} \quad [0,X]\times(y_1,y_2]\times[0,N_1]. \end{aligned} \end{align}\tag{140}\] By the definition of \(F\) in 133 , using 129 and 130 , we have, for \(\varepsilon\ll d_0^2\) and \(A\geq\frac{ CC_1}{d_0^2}\) that \[\begin{align} \label{Feq}\begin{aligned} &\partial_{x}F+(1+\tilde{q})\partial_{y}F -\frac{1}{u_{n-1}}\partial_{z} (\frac{u_n}{u_{n-1}}\partial_{z}F) +(-\tilde{b}-\frac{\partial_zu_{n} }{u_{n-1}^2})\partial_zF+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}F+c_1F \\ \leq &-\varepsilon^2d_0^2(\frac{2\lambda}{u_{n-1}^3}+\frac{A}{6}c_2)\phi_{1,\frac{\alpha}{2}} -(1+\tilde{q})\partial_{y}(K\partial_zu_n+(2-k_1)\varepsilon^2 e^{-\frac{A}{x+1}}C_1) \\&-[P_1 \phi_{1,\frac{\alpha}{2}}-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_z\phi_{1,\frac{\alpha}{2}}+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}\phi_{1,\frac{\alpha}{2}}+c_1\phi_{1,\frac{\alpha}{2}}](\frac{\varepsilon^2}{4}d_0^2-\gamma (y-y_*)) \\&-2\varepsilon^2 e^{-\frac{A}{x+1}}C_1(1-\zeta(y))(\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}+c_1) +2\varepsilon^2 e^{-\frac{A}{x+1}}C_1(1+\tilde{q})\partial_y \zeta \\&-2\varepsilon^2 e^{-\frac{A}{x+1}}C_1(1-\zeta(y))\frac{A}{(x+1)^2} +(1+\tilde{q})\gamma \phi_{1,\frac{\alpha}{2}} \\ \leq &-\varepsilon^2d_0^2(\frac{\lambda}{u_{n-1}^3}+\frac{A}{7}c_2)\phi_{1,\frac{\alpha}{2}} -(1+\tilde{q})\partial_{y}(K\partial_zu_n+(2-k_1)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1) \\& +(2-k_0)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1\partial_y \zeta, \end{aligned} \end{align}\tag{141}\] where \(d_0\) is defined in 56 which is independent of \(\varepsilon\) and \(\gamma\ll \varepsilon^3 e^{-A}\epsilon_0\) by 117 . Note that \[\begin{align} -2\varepsilon^2e^{-\frac{A}{x+1}} C_1(1-\zeta(y))\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}\leq 0, \end{align}\] and \(\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}+c_1>0\) for \(z\) near \(0\). By 116 , we have \[\begin{align} \label{etaeta} \frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}+c_1\geq -C e^{-\frac{(z+{\epsilon_0})^2}{x+1}\mu}\quad \text{in }\quad \Omega\cap\{0\leq y\leq Y^*\}. \end{align}\tag{142}\] By the definition of \(y_*,\) we have \(F_+(x,y_*,z)=0.\) Then by 132 and 140 , i.e. \(\partial_y F_+\geq 0\) in \([y_*,y_1)\) and \(\partial_y F_+\leq 0\) in \((y_1,y_2],\) we have \[\begin{align} &\int_{y_*}^{y_2} (1+\tilde{q})\partial_{y}(K\partial_zu_n+(2-k_1)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1) F_+ dy\\ =&(1+\tilde{q})(K\partial_zu_n+(2-k_1)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1) F_+(y_2) -\int_{y_*}^{y_2} \partial_{y}\tilde{q}(K\partial_zu_n+(2-k_1)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1) F_+ dy\\&- \int_{y_*}^{y_1 } (1+\tilde{q})(K\partial_zu_n+(2-k_1)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1) \partial_{y}F_+ dy- \int_{y_1}^{y_2 } (1+\tilde{q})(K\partial_zu_n+(2-k_1)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1) \partial_{y}F_+ dy \\ \geq& -\int_{y_*}^{y_2} \varepsilon^4 C\phi_{1,\frac{\alpha}{2}} F_+dy- \int_{y_*}^{y_1 } (2-k_0)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1 \partial_{y}F_+ dy- \int_{y_1}^{y_2 } (2-k_2)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1\partial_{y}F_+ dy \\ \geq& -\int_{y_*}^{y_2} \varepsilon^4C \phi_{1,\frac{\alpha}{2}} F_+dy- (2-k_0)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1 F_+ (y_1)+(1-k_3)(2-k_2)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1\,d(x)\,(y_2 -y_1), \end{align}\] where \(k_0\), \(k_1\), \(k_2\), \(k_3\) and \(d(x)\) are defined in 131 , 132 , 136 and 137 respectively. Hence, by 138 , \[\begin{align} \label{dyfkdzu}\begin{aligned} &\int_{y_*}^{y_2} - \frac{A}{7}c_2d_0^2\varepsilon^2\phi_{1,\frac{\alpha}{2}} F_+-(1+\tilde{q})\partial_{y}F F_+dy \\&+\int_{y_*}^{y_2} [-(1+\tilde{q})\partial_{y}(K\partial_zu_n+(2-k_1)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1) +(2-k_0)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1\partial_y \zeta] F_+ dy \\ \leq & C\int_{y_*}^{y_2} F_+ ^2 dy,\end{aligned} \end{align}\tag{143}\] where we have used \[\begin{align} \int_{y_*}^{y_2} -(1+\tilde{q})\partial_{y}FF_+dy =-\frac{1}{2} (1+\tilde{q})F_+^2(y_2)+\int_{y_*}^{y_2} \frac{1}{2} \partial_{y}\tilde{q}F_+^2dy \leq C\int_{y_*}^{y_2} F_+^2dy, \end{align}\] and \[\begin{align} \int_{y_*}^{y_2} (2-k_0)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1\partial_y \zeta F_+ dy \leq &-(2-k_0)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1 \frac{1}{y_2-y_1}\int_{y_1}^{y_2}F_+ dy\\ \leq &-(2-k_0)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1 \frac{1}{y_2-y_1}\int_{y_1}^{y_2} (F_+(y_1)+\int_{y_1}^y\partial_yF_+ dy')dy \\ \leq &-(2-k_0)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1F_+(y_1)+(1+k_3)(2-k_0)\varepsilon^{2}e^{-\frac{A}{x+1}}C_1\,d(x)\, \frac{y_2-y_1}{2} . \end{align}\] Moreover, by straightforward calculation, we have \[\begin{align} \label{dz0n1}\begin{aligned} &\int_0^{N_1}[-\frac{1}{u_{n-1}}\partial_{z} (\frac{u_n}{u_{n-1}}\partial_{z}F) +(-\tilde{b}-\frac{\partial_zu_{n} }{u_{n-1}^2})\partial_zF+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}F+c_1F]F_+dz \\=& \int_0^\infty-\frac{1}{u_{n-1}}\partial_{z} (\frac{u_n}{u_{n-1}}\partial_{z}F) F_+ +\frac{1}{2}(-\tilde{b}-\frac{\partial_zu_{n} }{u_{n-1}^2})\partial_zF_+^2+(\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}+c_1)F_+^2dz \\ =&\int_0^\infty \frac{u_n}{u_{n-1}^2}(\partial_{z}F_+)^2dz-\int_0^\infty \frac{\partial_zu_{n-1}u_n}{2u_{n-1}^3}\partial_{z}(F_+^2)dz \\ &+ \int_0^\infty\frac{1}{2}\partial_z(\tilde{b}+\frac{\partial_zu_{n} }{u_{n-1}^2})F_+^2+(\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}+c_1)F_+^2dz \\ \geq & -C_{{\epsilon_0}}\int_0^{N_1}F_+^2dz, \end{aligned} \end{align}\tag{144}\] where \(C_{{\epsilon_0}}\) is a positive constant depending on \({\epsilon_0}\) and \(N_1\) is defined in 124 . Then combining 143 and 144 , we have \[\begin{align} \partial_x\int_{y_*}^{y_2} \int_0^{N_1} F_+^2dzdy\leq C_{{\epsilon_0}} \int_{y_*}^{y_2} \int_0^{N_1} F_+^2dzdy \quad \text{in} \quad [0,X]. \end{align}\] Therefore, we have \(\int_{y_*}^{y_2} \int_0^{N_1} F_+^2dydz=0\) in \([0,X]\) because \(F_+(0,y,z)=0\) by Grownwall inequality. Therefore, we prove the goal 119 and 120 . We can prove the other direction \[\begin{align} -( \nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\tau_1}^2\bar{u})-\frac{\varepsilon^2}{2}d_0^2\phi_{1,\frac{\alpha}{2}} \leq 0 \quad \text{in} \quad[0,X]\times[0,Y^*]\times[0,+\infty), \end{align}\]in a same way. Then we complete the proof of the theorem. ◻

Proof. Step 1 First, we estimate \(|\nabla_{\eta}^2u_n-\nabla_{\tau_1}^2\bar{u}|.\)

Compared to the proof of Theorem 12, we only need to consider the term \[\begin{align} \label{03-26-30-1} \nabla_{\eta}(\frac{1}{u_{n-1}}\partial_z(\frac{\nabla_{\eta}\tilde{q}}{1+\tilde{q}}\frac{u_{n}}{u_{n-1}}\partial_z u_n)) \end{align}\tag{145}\] coming from \(\nabla_{\eta}f_1\) when we estimate \(|\nabla_{\eta}^2u_n-\nabla_{\tau_1}^2\bar{u}|,\) where \(f_1\) is defined in 98 . Since \[|\nabla_{\eta}(\frac{1}{u_{n-1}}\partial_z(\frac{\nabla_{\eta}\tilde{q}}{1+\tilde{q}}\frac{u_{n}}{u_{n-1}}\partial_z u_n)) -(\nabla_{\eta}\partial_z\nabla_{\eta}\tilde{q})\frac{1}{1+\tilde{q}}\frac{u_{n}}{u_{n-1}^2}\partial_z u_n |\leq \frac{C\phi_{1,0}}{u_{n-1}^{3-\frac{3\alpha}{2}}}.\] We will handle the term \((\nabla_{\eta}\partial_z\nabla_{\eta}\tilde{q})\frac{1}{1+\tilde{q}}\frac{u_{n}}{u_{n-1}^2}\partial_z u_n\).

We consider this new structural term because by definition of \(\nabla_\eta\), it contains the term \[\begin{align} \label{26-03-30-1} \frac{ \partial_y\partial_z\nabla_{\eta}\tilde{q}}{1+\tilde{q}}\frac{u_{n}}{u_{n-1}^2}\partial_z u_n=\partial_y(\frac{\partial_z\nabla_{\eta}\tilde{q}}{1+\tilde{q}}\frac{u_{n}}{u_{n-1}^2}\partial_z u_n)+Remainder. \end{align}\tag{146}\] The order of \(\partial_y\partial_z\nabla_{\eta}\tilde{q}\) is one order higher than the orders of derivatives in the induction assumption. We will handle it similarly as we handled the term \(\partial_y(K\partial_z u_n)\) in Theorem 12. However \[|\frac{\partial_z\nabla_{\eta}\tilde{q}}{1+\tilde{q}}\frac{u_{n}}{u_{n-1}^2}\partial_z u_n|\leq \frac{C}{u_{n-1}^2}\phi_{1,0}.\] An extra multiplier \(\frac{C}{u_{n-1}^2}\) increasing the growth rate near \(z=0\) should be taken into account. Then we modify the auxiliary function \(F\) where an extra \(u_n^2\) was multiplied to \(f\). i.e. Take \(y_2-y_*\) small enough such that \[\begin{align} \label{y2u2} \gamma (1+\|\phi_{1,\frac{\alpha}{2}}\|_{L^\infty})(y_2-y_*)\leq \gamma u^2_{n}, \end{align}\tag{147}\] and \[\begin{align} \label{26-06-14-1} f\leq \epsilon_0\quad\text{in}\quad[0,X]\times[y_*,y_2]\times[0,N_1], \end{align}\tag{148}\] where we recall the definition of \(y_*\) defined in 121 . Set \[f=\nabla_{\eta}^2 u_n-\nabla_{\tau_1}^2 \bar{u}-\frac{\varepsilon^2}{2}d_0^2\phi_{1,\frac{\alpha}{2}}-\gamma\quad \text{in}\quad [0,X]\times[0,Y]\times[0,+\infty),\] and in \([y_*,y_2]\), \[\begin{align} \label{26-03-29-2} \begin{aligned} F:=&u_n^2f -2\varepsilon^{2}e^{-\frac{A}{x+1}}C_1+2 \varepsilon^{2}e^{-\frac{A}{x+1}}C_1\zeta(y) +\gamma \phi_{1,\frac{\alpha}{2}}(y-y_*), \end{aligned} \end{align}\tag{149}\] where \(\zeta(y)\) is defined in 134 . Then \[\begin{align} &P_1 F-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_zF+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}F+c_1F\\=& \partial_y(Ku_n^2\partial_z u_n)+ \partial_y(\frac{\partial_z\nabla_{\eta}\tilde{q}}{1+\tilde{q}}\frac{u_{n}^3}{u_{n-1}^2}\partial_z u_n)+\cdots, \end{align}\] where the second term on the right hand side corresponds to 146 multiplied by \(u_n^2\) and the first two terms on the right hand side can be estimated like \(\partial_y (K\partial_z u_n)\) in the previous proof, since \[|Ku_n^2\partial_z u_n|+|\frac{\partial_z\nabla_{\eta}\tilde{q}}{1+\tilde{q}}\frac{u_{n}^3}{u_{n-1}^2}\partial_z u_n|\leq \varepsilon^2 e^{-\frac{A}{x+1}}C.\] Next we handle the new terms appearing in the equation of \(F\) when we use \(u_n^2f.\)

Step 1.1 Equation of \(f\)

Comparing 98 with 99 , similar to 128 we have \[\begin{align} \label{26-03-30-4}\begin{aligned} &P_1 f-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_zf+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}f+c_1f \\ =&\frac{1}{u_n^2}\partial_{y}(Ku_n^2\partial_zu_n)+ \frac{1}{u_n^2}\partial_y( \frac{\partial_z\nabla_{\eta}\tilde{q}}{1+\tilde{q}}\frac{u_{n}^3}{u_{n-1}^2}\partial_z u_n) -f_{11}^u-\gamma(\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}+c_1) \\&-[P_1 \phi_{1,\frac{\alpha}{2}}-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_z\phi_{1,\frac{\alpha}{2}}+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}\phi_{1,\frac{\alpha}{2}}+c_1\phi_{1,\frac{\alpha}{2}}]\frac{\varepsilon^2}{2}d_0^2 ,\end{aligned} \end{align}\tag{150}\] with \[\begin{align} |c_1|\leq \frac{C \phi_{1,0}}{u_{n-1}^2}, \quad |f_{11}^{u}|\leq C\varepsilon^2\phi_{1,0}+\frac{\varepsilon^2C\phi_{1,0}}{u_{n-1}^{3-\alpha}}. \end{align}\] Here we have used \[\begin{align} \frac{1}{u_n^2}\partial_{y}(K\partial_zu_n)u_n^2=& \frac{1}{u_n^2}\partial_{y}(Ku_n^2\partial_zu_n)-\frac{1}{u_n^2}K\partial_{y}u_n^2\partial_zu_n,\\ \frac{1}{u_n^2}\frac{\partial_y\partial_z\nabla_{\eta}\tilde{q}}{1+\tilde{q}}\frac{u_{n}^3}{u_{n-1}^2}\partial_z u_n= & \frac{1}{u_n^2}\partial_y( \frac{\partial_z\nabla_{\eta}\tilde{q}}{1+\tilde{q}}\frac{u_{n}^3}{u_{n-1}^2}\partial_z u_n)-\frac{1}{u_n^2}\partial_z\nabla_{\eta}\tilde{q}\partial_y ( \frac{u_{n}^3}{(1+\tilde{q})u_{n-1}^2}\partial_z u_n), \\ \nabla_{\eta}\partial_z\nabla_{\eta}\tilde{q}=&\partial_y\partial_z\nabla_{\eta}\tilde{q} -\frac{\int_0^z \partial_yv_{n-1}dz'}{v_{n-1}}\partial_z^2\nabla_{\eta}\tilde{q}, \end{align}\]and \[\begin{align} \label{sm3} |\frac{\int_0^z \partial_yv_{n-1}dz'}{v_{n-1}}\partial_z^2\nabla_{\eta}\tilde{q}|\leq \frac{C}{u_{n-1}^{1-\alpha}}\phi_{1,0} \quad\text{in}\quad \Omega\cap\{0\leq y\leq Y^*\}, \end{align}\tag{151}\] which imply \[|(\nabla_{\eta}\partial_z\nabla_{\eta}\tilde{q}-\partial_y\partial_z\nabla_{\eta}\tilde{q})\frac{1}{1+\tilde{q}}\frac{u_{n}}{u_{n-1}^2}\partial_z u_n |\leq \frac{C}{u_{n-1}^{2-\alpha}}\phi_{1,0},\] because \(|\frac{1}{1+\tilde{q}}\frac{u_{n}}{u_{n-1}^2}\partial_z u_n|\leq \frac{C}{u_{n-1}}.\) 151 comes from \[\begin{align} \label{26-03-17-1} |\partial_z^2\nabla_{\eta}\tilde{q}|= |\partial_z^2(\partial_y\tilde{q}-\frac{\int_0^z \partial_yv_{n-1}dz'}{v_{n-1}}\partial_z\tilde{q})| \leq \frac{C}{u_{n-1}^{2-\alpha}}\phi_{1,0}, \end{align}\tag{152}\] which can be derived from the equations of \(u_{n-1},\) \(v_{n-1}\) and the induction assumption in 44 .

Step 1.2 Equation of \(g=u_n^2f\)

By definition of \(P_1\) in 68 , we have \(g=u_{n}^2f\) satisfies

\[\begin{align} \begin{aligned} P_1g-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_zg =&f[P_1u_{n}^2 -\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_zu_n^2] +u_n^2[P_1f-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_zf] \\& -2\frac{u_{n}}{u_{n-1}^2}\partial_{z}u_n^2\partial_{z}f\\ =&f[\frac{-2u_{n}(\partial_zu_{n} )^2 }{u_{n-1}^2}-\frac{2u_n(\partial_zu_{n})^2 }{u_{n-1}^2}] +u_n^2[P_1f-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_zf] \\& -4\frac{u_{n}^2}{u_{n-1}^2}\partial_{z}u_n\partial_{z}f\\ =&u_n^2[P_1f-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_zf] -fI_1 -\partial_{z}f I_2,\end{aligned} \end{align}\] with \[0\leq I_1\leq C\frac{u_n}{u_{n-1}^2}\leq C\frac{1}{u_{n-1}},\quad 0\leq I_2\leq C,\] where we have used by 76 , \[P_1 u_n^2=\frac{-2u_{n}(\partial_zu_{n} )^2 }{u_{n-1}^2} .\] Then \[\begin{align} \label{26-04-24-1}\begin{aligned} &P_1 g-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_zg+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}g+c_1g\\ \leq &u_n^2[P_1f-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_zf+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}f+c_1f] \\& +I_3\partial_zF+I_4 F +Ce^{-\frac{(z+\epsilon_0)^2}{x+1}}+Cu_{n-1}^2\phi_{1,\frac{\alpha}{2}}(y-y_*) ,\end{aligned} \end{align}\tag{153}\] with \(|I_3|+|\partial_zI_3|+|I_4|\leq C_{\epsilon_0}\) Note that \[P_1f-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_zf+\frac{\partial_zu_{n} \partial_zu_{n-1} }{u_{n-1}^3}f+c_1f\] follows from the equation of \(f\) in 150 . Take \[\gamma\leq e^{-A}\epsilon_0^5,\] which implies \(\gamma\leq e^{-A}u_{n-1}^5\ll u_{n-1}^5.\) Then by the definition of \(F\) in 149 , the following inequalities hold in \([y_*,y_2]\), \[\begin{align} f\geq &\frac{1}{u_n^2}F-\frac{1}{u_n^2}\gamma \phi_{1,\frac{\alpha}{2}}(y-y_*) \geq \frac{1}{u_n^2}F-u_{n-1}^3\phi_{1,\frac{\alpha}{2}}(y-y_*), \\ \frac{u_n^2}{u_{n-1}^2}\partial_z f\geq &\frac{1}{u_{n-1}^2}[\partial_zF-f\partial_zu_n^2-\gamma \partial_z\phi_{1,\frac{\alpha}{2}}(y-y_*)]\\ \geq &\frac{1}{u_{n-1}^2}\partial_zF-f\frac{2u_n\partial_zu_n}{u_{n-1}^2}-u_{n-1}^2\phi_{1,\frac{\alpha}{2}}(y-y_*) \\ \geq&\frac{1}{u_{n-1}^2}\partial_zF-\frac{\epsilon_0C}{u_{n-1}}e^{-\frac{(z+\epsilon_0)^2}{x+1}}-u_{n-1}^2\phi_{1,\frac{\alpha}{2}}(y-y_*) \\ \geq&\frac{1}{u_{n-1}^2}\partial_zF-Ce^{-\frac{(z+\epsilon_0)^2}{x+1}}-u_{n-1}^2\phi_{1,\frac{\alpha}{2}}(y-y_*) \end{align}\] where we have used \[-\partial_z\phi_{1,\frac{\alpha}{2}}\geq-\frac{\alpha}{2(z+{\epsilon_0})} \phi_{1,\frac{\alpha}{2}}\geq -\frac{C}{u_{n-1}}\phi_{1,\frac{\alpha}{2}},\] which can be derived by the definition of \(\phi_{1, \frac{\alpha}{2}}\) in 45 and the following calculation, \[\partial_z\phi_{1,\frac{\alpha}{2}}= \left\{ \begin{align} \frac{\alpha}{2(z+{\epsilon_0})} \phi_{1,\frac{\alpha}{2}},\quad & 0\leq -\frac{z+{\epsilon_0}}{\sqrt{x+1}}\leq \delta,\\ 0,\quad & \delta\leq \frac{z+{\epsilon_0}}{\sqrt{x+1}}\leq N,\\ -\frac{2(z+{\epsilon_0})}{x+1}\mu\phi_{1,\frac{\alpha}{2}} ,\quad\quad & \frac{z+{\epsilon_0}}{\sqrt{x+1}}\geq N . \end{align}\right.\\\]

Then the proof proceeds similarly as the proof of Theorem 12.

Step 2 Next, we prove \(|\nabla_{\xi}\nabla_{\eta,\xi}u_n-\nabla_{\tau_1}^2\bar{u}|\leq \frac{\varepsilon^2}{2}d_0^2\phi_{1,\frac{\alpha}{2}}\).

Recall 98 and 99 for the equations of \(\nabla_{\eta} u_n\) and \(\nabla_{\xi } u_n\). Then the key term \((1+\tilde{q})\partial_{y}(K\partial_zu_n)\) in 113 is replaced by \(\partial_x(\mp K\partial_z u_n )\) and \((1+\tilde{q})\partial_x(\pm K\partial_z u_n)\) when we estimate \(\pm(\nabla_{\xi}\nabla_{\eta,\xi}u_n-\nabla_{\tau_1}^2\bar{u})\). The proof is similar to the proof of Theorem 12 and Step 1. For example, we will estimate \(\nabla_{\xi}^2u_n-\nabla_{\tau_1}^2\bar{u}\) where \((1+\tilde{q})\partial_{y}(K\partial_zu_n)\) is replaced by \((1+\tilde{q})\partial_x( K\partial_z u_n)\). Set \[\begin{align} f= ( \nabla_{\xi}^2 u_n-\nabla_{\tau_1}^2\bar{u})-\frac{\varepsilon^2}{2}d_0^2\phi_{1,\frac{\alpha}{2}}-\gamma \quad \text{in} \quad[0,X]\times[0,Y]\times[0,+\infty). \end{align}\] As in the proof of Theorem 12, we apply proof by contradiction. Instead of 121 , set \[x_*:=\max \{x_0\in[0,X]|f(x,y,z)\leq 0,\,\,\{0\leq x\leq x_0\} \times[0,Y^*]\times[0,+\infty)\}<X.\] By the boundary condition, for some positive constant \(y_0<Y^*,\) \[\begin{align} \label{d32025} f<0\quad \text{in}\quad[0,X]\times[0,y_0]\times[0,N_1]\cup[0,X]\times[0,Y^*]\times[0,z_0]. \end{align}\tag{154}\] Similar estimates as in the proof of Theorem 12 hold for the sign of \(f\) and \(\partial_x f\) on \[D_{x_*}=\{x=x_*\}\times[y_0,Y^*]\times[z_0,N_1].\] Take a constant \(x_2\) such that \(x_2-x_*\) is sufficiently small and \(0<x_2-x_*\ll\min\{1,X-x_*\}\). Then by the continuity, for some domains \(D_1\) and \(D_2\) with \(D_1\cup D_2=[0,Y^*]\times[0,N_1]\), \[\begin{align} \label{26-03-15-3}\begin{aligned} f+\gamma \phi_{1,\frac{\alpha}{2}}(x-x_*) <&0 \quad \quad\quad\quad\quad \text{in} \quad [x_*,x_2]\times D_1 ,\\ \partial_x (f+\gamma \phi_{1,\frac{\alpha}{2}}(x-x_*))\geq&0 \quad\quad\quad\quad\quad \text{in} \quad [x_*,x_2]\times D_2,\end{aligned} \end{align}\tag{155}\] with noting \(\partial_x \phi_{1,\frac{\alpha}{2}}\geq 0\). Furthermore, take \(x_2-x_*\) sufficiently small such that \[\begin{align} \label{2026-03-15-1} |\frac{ e^{-\frac{A}{x_2+1}}}{ e^{-\frac{A}{x_*+1}}}-1|\ll 1, \end{align}\tag{156}\] which implies \[\begin{align} e^{-\frac{A}{x_2+1}}e^{-\frac{(z+{\epsilon_0})^2}{x+1}\mu} \leq C\phi_{1,0}\quad \text{in}\quad[x_*,x_2]\times[0,Y^*]\times[0,+\infty), \end{align}\] and instead of 135 , \[\begin{align} \label{26-03-15-2} 2 \varepsilon^{2}e^{-A}C_1\frac{1}{x_2-x_1}\gg\max_{[0,X]\times[0,Y^*]\times[0,N_1]} |\partial_x f|+\gamma |\partial_x\big(\phi_{1,\frac{\alpha}{2}}(x-x_*)\big)|. \end{align}\tag{157}\] Next, instead of 133 , set\[F= f -2\varepsilon^{2}e^{-\frac{A}{x_2+1}}C_1(1-\zeta) +\gamma \phi_{1,\frac{\alpha}{2}}(x-x_*), \quad\text{in} \quad[x_*,x_2]\times[0,Y^*]\times[0,+\infty),\] where \[\zeta(x)= \left\{ \begin{align} 1,\quad & x_*\leq x\leq x_1,\\ 1-\frac{x-x_1}{x_2-x_1},\quad & x_1< x\leq x_2, \end{align}\right.\\\] By 157 , it holds \[\begin{align} \label{26-03-15-4} 1-k_3 \leq \frac{ -\partial_x F}{d}\leq 1+k_3\quad\text{in} \quad(x_1,x_2], \end{align}\tag{158}\] with \[\begin{align} d= \frac{2 \varepsilon^{2}e^{-\frac{A}{x_2+1}}C_1}{x_2-x_1}, \end{align}\] which corresponds to 136 . Then by \(1-\zeta\geq 0,\) 155 and 158 , we have \[\begin{align} \begin{aligned} \partial_x F_+\geq &0\quad \text{in} \quad [x_*,x_1)\times[0,Y^*]\times[0,N_1],\\ \partial_x F_+\leq &0\quad \text{in} \quad (x_1,x_2]\times[0,Y^*]\times[0,N_1]. \end{aligned} \end{align}\] Then instead of 143 , by integrating in \(x\) and using a similar argument as in the proof of Theorem 12, we have \[\begin{align} \begin{aligned} &\int_{x_*}^{x_2} - \frac{A}{7}c_2d_0^2\varepsilon^2\phi_{1,\frac{\alpha}{2}} F_+-\partial_{x}F F_+dx \\&+\int_{x_*}^{x_2} [-(1+\tilde{q})\partial_{x}(K\partial_zu_n+(2-k_1)\varepsilon^{2}e^{-\frac{A}{x_2+1}}C_1) +(2-k_0)\varepsilon^{2}e^{-\frac{A}{x_2+1}}C_1\partial_x \zeta] F_+ dx \\ \leq & C\int_{x_*}^{x_2} F_+ ^2 dx.\end{aligned} \end{align}\] Then we have \[\begin{align} \int_{x_*}^{x_2}\int_0^{N_1}(1+\tilde{q})\partial_{y}F F_+ dzdx\leq C_{\epsilon_0}\int_{x_*}^{x_2}\int_0^{N_1} F_+ ^2 dzdx \end{align}\] similar to 144 . Thus \[\begin{align} \begin{aligned} \partial_{y} \int_{x_*}^{x_2}\int_0^{N_1}(1+\tilde{q})F_+^2dzdx = & \int_{x_*}^{x_2}\int_0^{N_1}(1+\tilde{q})\partial_{y}F_+^2 +\partial_{y}\tilde{q}\,F_+^2dzdx \\ = &\int_{x_*}^{x_2}\int_0^{N_1}2(1+\tilde{q})\partial_{y}F F_+ +\partial_{y}\tilde{q}\,F_+^2dzdx \\ \leq & C_{\epsilon_0}\int_{x_*}^{x_2}\int_0^{N_1} F_+ ^2 dzdx \\ \leq & C_{\epsilon_0}\int_{x_*}^{x_2}\int_0^{N_1} (1+\tilde{q})F_+ ^2 dzdx\quad\text{in}\quad [0,Y^*].\end{aligned} \end{align}\] Hence, we have the conclusion \(\int_{x_*}^{x_2}\int_0^{N_1}(1+\tilde{q})F_+^2dzdx=0\) in \([0,Y^*]\) because \(F_+(x,0,z)=0\) by Grownwall inequality. This leads to a contradiction to the definition of \(x_*\) and the proof of the theorem is complete. ◻

Proof. We will prove \[\begin{align} \label{1106} |\nabla_{\eta}\nabla_{\xi}u_n-\nabla_{\eta}\nabla_{\xi}\bar{u}|\leq \frac{\varepsilon^2}{2}d_0^2\phi_{1,\frac{\alpha}{2}} \quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}, \end{align}\tag{159}\] and the other inequalities can be derived similarly.

By 113 , \[\begin{align} \label{26-03-29-1}\begin{aligned} 0=&P_1( \nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\eta}\nabla_{\xi}\bar{u})-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_z( \nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\eta}\nabla_{\xi}\bar{u})+\tilde{c}( \nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\eta}\nabla_{\xi}\bar{u})\\&+(1+\tilde{q})\partial_{y}(K\partial_zu_n) +F^{new},\end{aligned} \end{align}\tag{160}\] where \[F^{new}:=-[P_1( \nabla_{\tau_1}^2\bar{u} -\nabla_{\eta}\nabla_{\xi}\bar{u})-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_z( \nabla_{\tau_1}^2\bar{u}-\nabla_{\eta}\nabla_{\xi}\bar{u})+\tilde{c}( \nabla_{\tau_1}^2\bar{u}-\nabla_{\eta}\nabla_{\xi}\bar{u})]+f_{12}^u,\] where \(f_{12}^u\) and \(\tilde{c}\) are the functions defined in 114 and 115 respectively.

Step 1 Decompose \(F^{new}\) into several structural components

Based on the following claims, we have \[\begin{align} \label{26-03-30-5} F^{new}:=\partial_yQ +I+f_{12}^u, \end{align}\tag{161}\] with \[\begin{align} \label{26-03-30-6} |Q|+ |I|\leq \frac{C}{u_{n-1}^2}\phi_{1,0}\quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}. \end{align}\tag{162}\]

Claim 1. The first two terms for \(P_1( \nabla_{\tau_1}^2\bar{u} -\nabla_{\eta}\nabla_{\xi}\bar{u})\) with \(P_1\) defined in 68 satisfy\[\begin{align} & \partial_x ( \nabla_{\tau_1}^2\bar{u} -\nabla_{\eta}\nabla_{\xi}\bar{u}) +(1+\tilde{q})\partial_{y}( \nabla_{\tau_1}^2\bar{u} -\nabla_{\eta}\nabla_{\xi}\bar{u}) \\=&\partial_{y}( Q_1)\partial_z \bar{u}+(1+\tilde{q})\partial_{y}( Q_2)\partial_z \bar{u}+\tilde{I}_1 \\=&\partial_{y}( Q_3)+\tilde{I}_2, \end{align}\] where \(Q_3=(Q_1+(1+\tilde{q}) Q_2)\partial_z \bar{u}\), \[\begin{align} Q_1&=-\frac{\int_0^z \partial_{x}\partial_x\bar{u}dz'}{\bar{u}}+\frac{\int_0^z \partial_{x}\partial_xu_{n-1}dz'}{u_{n-1}},\\ Q_2&=-\frac{\int_0^z \partial_{y}\partial_x\bar{u}dz'}{\bar{u}}+\frac{\int_0^z \partial_{y}\partial_xu_{n-1}dz'}{u_{n-1}}, \end{align}\] and \(\tilde{I}_1, \,\tilde{I}_2\) do not contain third-order tangential derivatives which are one order higher than the orders of derivatives in the induction assumption. In particular, by 44 , \(Q_3\) satisfies \[\begin{align} |Q_3|\leq 2\varepsilon^2 e^{-\frac{A}{x+1}}C,\quad|\tilde{I}_2|\leq C\frac{1}{u_{n}^2}\phi_{1,0} \quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}, \end{align}\] for some constant \(C\) independent of \(\varepsilon.\)

Claim 2. The fourth and fifth terms of \(P_1( \nabla_{\tau_1}^2\bar{u} -\nabla_{\eta}\nabla_{\xi}\bar{u})\) are \[\begin{align} \begin{aligned} &-\frac{u_{n}}{u_{n-1}^2}\partial_{z}^2(( \nabla_{\tau_1}^2\bar{u} -\nabla_{\eta}\nabla_{\xi}\bar{u}))-\frac{1}{u_{n-1}}\partial_z(\frac{u_{n}}{u_{n-1}})\,\partial_{z} ( \nabla_{\tau_1}^2\bar{u} -\nabla_{\eta}\nabla_{\xi}\bar{u})\\ =& \partial_y( Q_6)+I_6 ,\end{aligned} \end{align}\] with \[\begin{align} |Q_6|+ |I_6|\leq \frac{C}{u_{n-1}^2}\phi_{1,0}\quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}. \end{align}\]

Claim 3. \[\begin{align} -\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_z( \nabla_{\tau_1}^2\bar{u}-\nabla_{\eta}\nabla_{\xi}\bar{u})+\tilde{c}( \nabla_{\tau_1}^2\bar{u}-\nabla_{\eta}\nabla_{\xi}\bar{u}) = \partial_y( Q_9)+I_9, \end{align}\]with \[\begin{align} |Q_9|+ |I_9|\leq \frac{C}{u_{n-1}^2}\phi_{1,0}\quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}. \end{align}\]

Proof of the claims:

Growth rates of \(\bar{u}\) and its derivatives are given in subsection 7.3.

For Claim 1, by the definitions, \[\begin{align} \nabla_{\eta}=\partial_y-\frac{\int_0^z \partial_yv_{n-1}dz'}{v_{n-1}}\partial_z, \quad \nabla_{\xi}&=\partial_x-\frac{\int_0^z \partial_xu_{n-1}dz'}{u_{n-1}}\partial_z. \end{align}\]Since \[\begin{align} \begin{aligned} \partial_{x,y}\partial_{y}(-\frac{\int_0^z \partial_xu_{n-1}dz'}{u_{n-1}})\partial_z \bar{u}=& \partial_{y}\partial_{x,y}(-\frac{\int_0^z \partial_xu_{n-1}dz'}{u_{n-1}})\partial_z \bar{u} \\ :=& \partial_{y}( -\frac{\int_0^z \partial_{x,y}\partial_xu_{n-1}dz'}{u_{n-1}})\partial_z \bar{u}+I_0,\end{aligned} \end{align}\] we have \[\begin{align} \label{26-03-15-5-1-1}\begin{aligned} \partial_{x,y}\nabla_{\eta}\nabla_{\xi}\bar{u}:=& \partial_{x,y} \partial_{y}\partial_x\bar{u}+\partial_{y}( -\frac{\int_0^z \partial_{x,y}\partial_xu_{n-1}dz'}{u_{n-1}})\partial_z \bar{u}+I_1 ,\end{aligned} \end{align}\tag{163}\] where \(I_0\) and \(I_1\) do not contain the third-order tangential derivatives \(\partial_{x,y}^3u_{n-1}.\) By the definitions, \(\nabla_{\tau_1} = \partial_x-\frac{\int_0^z \partial_{x}\bar{u}dz'}{\bar{u}}\partial_z\) and \(\nabla_{\tau_2} = \partial_y-\frac{\int_0^z \partial_{y}\bar{u}dz'}{\bar{u}}\partial_z.\) By the symmetry of \(\bar{u},\) we have \[\nabla_{\tau_1}^2\bar{u}=\nabla_{\tau_2}\nabla_{\tau_1}\bar{u}.\] Hence, similarly to 163 , it holds \[\begin{align} \partial_{x,y} \nabla_{\tau_1}^2\bar{u} := \partial_{x,y}\partial_{y}\partial_x\bar{u}+\partial_{y}( -\frac{\int_0^z \partial_{x,y}\partial_x\bar{u}dz'}{\bar{u}})\partial_z \bar{u}+I_2, \end{align}\] where \(I_2\) does not contain the third-order tangential derivatives \(\partial_{x,y}^3\bar{u}.\) In summary, \[\begin{align} \partial_{x,y} ( \nabla_{\tau_1}^2\bar{u} -\nabla_{\eta}\nabla_{\xi}\bar{u}):= \partial_{y}( -\frac{\int_0^z \partial_{x,y}\partial_x\bar{u}dz'}{\bar{u}}+\frac{\int_0^z \partial_{x,y}\partial_xu_{n-1}dz'}{u_{n-1}})\partial_z \bar{u}+I_3, \end{align}\]where \(I_3\) does not contain the third-order tangential derivatives. Then we can prove claim 1 by straight forward calculation and growth estimates.

Similarly, for Claim 2, note that \[\begin{align} \begin{aligned} \partial_{z}\partial_{y}(-\frac{\int_0^z \partial_xu_{n-1}dz'}{u_{n-1}})\partial_z \bar{u}=& \partial_{y}\partial_{z}(-\frac{\int_0^z \partial_xu_{n-1}dz'}{u_{n-1}})\partial_z \bar{u} \\ =& \partial_{y}( -\frac{ \partial_xu_{n-1}}{u_{n-1}}+\frac{\int_0^z \partial_xu_{n-1}dz'}{u_{n-1}^2}\partial_zu_{n-1})\partial_z \bar{u} \\ =& \partial_{y}Q_4-( -\frac{ \partial_xu_{n-1}}{u_{n-1}}+\frac{\int_0^z \partial_xu_{n-1}dz'}{u_{n-1}^2}\partial_zu_{n-1})\partial_z\partial_{y} \bar{u}, \\ \partial_{z}^2\partial_{y}(-\frac{\int_0^z \partial_xu_{n-1}dz'}{u_{n-1}})\partial_z \bar{u} =& \partial_{y}\partial_z( -\frac{ \partial_xu_{n-1}}{u_{n-1}}+\frac{\int_0^z \partial_xu_{n-1}dz'}{u_{n-1}^2}\partial_zu_{n-1})\partial_z \bar{u}\\ =&\partial_{y} Q_5-\partial_z( -\frac{ \partial_xu_{n-1}}{u_{n-1}}+\frac{\int_0^z \partial_xu_{n-1}dz'}{u_{n-1}^2}\partial_zu_{n-1})\partial_z \partial_{y}\bar{u},\end{aligned} \end{align}\] where \[\begin{align} Q_4=& ( -\frac{ \partial_xu_{n-1}}{u_{n-1}}+\frac{\int_0^z \partial_xu_{n-1}dz'}{u_{n-1}^2}\partial_zu_{n-1})\partial_z \bar{u},\\ Q_5=&\partial_z( -\frac{ \partial_xu_{n-1}}{u_{n-1}}+\frac{\int_0^z \partial_xu_{n-1}dz'}{u_{n-1}^2}\partial_zu_{n-1})\partial_z \bar{u}. \end{align}\] Here, the order of differentiation in \(\partial_{y}\partial_z \partial_xu_{n-1}\) is one order higher than the order of derivatives in the induction assumption. Then by straight forward calculation, we have \[\begin{align} \label{26-03-15-5-1}\begin{aligned} \partial_{z}( \nabla_{\tau_1}^2\bar{u}-\nabla_{\eta}\nabla_{\xi}\bar{u}):=&\partial_y Q_4+I_4,\\ \partial_{z}^2( \nabla_{\tau_1}^2\bar{u}-\nabla_{\eta}\nabla_{\xi}\bar{u}):=&\partial_y Q_5+I_5,\end{aligned} \end{align}\tag{164}\] with \[\begin{align} \label{26-03-30-2} |I_4|+u_{n-1}|I_5| \leq C\phi_{1,0},\quad|Q_4|\leq C\phi_{1,0},\quad |Q_5|\leq\frac{C}{u_{n-1}}\phi_{1,0}\quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}, \end{align}\tag{165}\] by the induction assumptions and in particular, by noting \[|\partial_y u_{n-1}|\leq Cu_{n-1}.\] Moreover, since \(|\partial_y Q_4|\leq\frac{C}{u_{n-1}}\phi_{1,0}\) by induction assumption, \[\begin{align} \label{26-03-30-3}\begin{aligned} u_{n-1}|\partial_{z}( \nabla_{\tau_1}^2\bar{u}-\nabla_{\eta}\nabla_{\xi}\bar{u})|\leq&C\phi_{1,0}, \quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}. \end{aligned} \end{align}\tag{166}\]

Then by 68 and 164 , the fourth and fifth terms of \(P_1( \nabla_{\tau_1}^2\bar{u} -\nabla_{\eta}\nabla_{\xi}\bar{u})\) are \[\begin{align} \begin{aligned} &-\frac{u_{n}}{u_{n-1}^2}\partial_{z}^2( \nabla_{\tau_1}^2\bar{u} -\nabla_{\eta}\nabla_{\xi}\bar{u})-\frac{1}{u_{n-1}}\partial_z(\frac{u_{n}}{u_{n-1}})\,\partial_{z} ( \nabla_{\tau_1}^2\bar{u} -\nabla_{\eta}\nabla_{\xi}\bar{u})\\ =& \partial_y (-\frac{u_{n}}{u_{n-1}^2}Q_5)-Q_5\partial_y (-\frac{u_{n}}{u_{n-1}^2})+\partial_y[(\frac{u_{n}}{u_{n-1}^3}\partial_zu_{n-1}-\frac{\partial_zu_{n}}{u_{n-1}^2})Q_4] \\&-Q_4\partial_y(\frac{u_{n}}{u_{n-1}^3}\partial_zu_{n-1}-\frac{\partial_zu_{n}}{u_{n-1}^2}) -\frac{u_{n}}{u_{n-1}^2}I_5+(\frac{u_{n}}{u_{n-1}^3}\partial_zu_{n-1}-\frac{\partial_zu_{n}}{u_{n-1}^2})I_4\\ =& \partial_y( Q_6)+I_6 ,\end{aligned} \end{align}\] with \[\begin{align} |Q_6|+ |I_6|\leq \frac{C}{u_{n-1}^2}\phi_{1,0}\quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}, \end{align}\] by the induction assumptions and in particular, by noting \(|\partial_y u_{n-1}|\leq Cu_{n-1}.\)

For Claim 3, by the same method in the proof of claim 2, we can decompose \(-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_z( \nabla_{\tau_1}^2\bar{u}-\nabla_{\eta}\nabla_{\xi}\bar{u})\) as follows. Now we decompose the term \(\tilde{c}( \nabla_{\tau_1}^2\bar{u}-\nabla_{\eta}\nabla_{\xi}\bar{u}).\)

Since \[\begin{align} \begin{aligned} \nabla_{\eta}\nabla_{\xi}\bar{u}=& \partial_{y}\partial_x\bar{u}+\partial_{y}( -\frac{\int_0^z \partial_xu_{n-1}dz'}{u_{n-1}}\partial_z \bar{u}) \\&-\frac{\int_0^z \partial_yv_{n-1}dz'}{v_{n-1}}\partial_z\partial_x\bar{u}-\frac{\int_0^z \partial_yv_{n-1}dz'}{v_{n-1}}\partial_z( -\frac{\int_0^z \partial_xu_{n-1}dz'}{u_{n-1}}\partial_z \bar{u}),\end{aligned} \end{align}\]it holds \[\nabla_{\tau_1}^2\bar{u}-\nabla_{\eta}\nabla_{\xi}\bar{u}=\partial_y( Q_7)+I_7,\] where \(Q_7= -\frac{\int_0^z \partial_x\bar{u}dz'}{\bar{u}}\partial_z \bar{u}+\frac{\int_0^z \partial_xu_{n-1}dz'}{u_{n-1}}\partial_z \bar{u}\) with \[\begin{align} |Q_7|+ | I_7|\leq&Cu_{n-1}\phi_{1,0} \quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}. \end{align}\]

Moreover, by the expression of \(\tilde{c}\) in 115 , the induction assumptions and in particular, by noting \(|\partial_y u_{n-1}|\leq Cu_{n-1},\) we have \[\begin{align} |\tilde{c}|+|\partial_y\tilde{c}|\leq \frac{C}{u_{n-1}^3}e^{-\frac{(z+{\epsilon_0})^2}{x+1}\mu}\quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}. \end{align}\] Then we have \[\begin{align} \begin{aligned} \tilde{c}( \nabla_{\tau_1}^2\bar{u}-\nabla_{\eta}\nabla_{\xi}\bar{u})=&\partial_y( Q_7\tilde{c})+I_7\tilde{c}-\partial_y\tilde{c}Q_7\\ =&\partial_y( Q_8)+I_8 ,\end{aligned} \end{align}\] with \[\begin{align} |Q_8|+ | I_8|\leq&\frac{C}{u_{n-1}^2}\phi_{1,0} \quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}. \end{align}\] In summary, all the three claims hold.

Step 2 Reduction to the case similar to 150

Next, by 160 and 161 162 , \(\nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\eta}\nabla_{\xi}\bar{u}\) satisfies the following equation \[\begin{align} \begin{aligned} 0 =&P_1( \nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\eta}\nabla_{\xi}\bar{u})-\frac{\partial_zu_{n} }{u_{n-1}^2}\partial_z( \nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\eta}\nabla_{\xi}\bar{u})+\tilde{c}( \nabla_{\eta}\nabla_{\xi} u_n-\nabla_{\eta}\nabla_{\xi}\bar{u})\\&+\frac{1}{u_n^2}\partial_{y}E +f_{12}^{new},\end{aligned} \end{align}\] with \[\begin{align} E=[(1+\tilde{q})K\partial_zu_n+Q]u_n^2, \end{align}\] and \[\begin{align} \label{26-03-15-7} |E|\leq 2\varepsilon^2 e^{-\frac{A}{x+1}}C_1 \quad \text{in} \quad \Omega\cap\{0\leq y\leq Y^*\}, \end{align}\tag{167}\] where \[\begin{align} f_{12}^{new}:=&I-\partial_{y}\tilde{q}\,K\partial_zu_n -2\frac{\partial_y u_n}{u_n}[(1+\tilde{q})K\partial_zu_n+Q] +f_{12}^u. \end{align}\] Then by \(|\partial_y u_n|\leq Cu_n\), 162 and 116 , \(f_{12}^{new}\) satisfies \[\begin{align} |f_{12}^{new}|\leq \frac{C}{u_{n-1}^{2}}\phi_{1,0} +\frac{\varepsilon^2C\phi_{1,0}}{u_{n-1}^{3-\alpha}}, \end{align}\] which can be controlled by the barrier function. Then we derive 159 by using the same method as for the term 146 through the estimation on \(F=u_n^2 f+remainder\) defined in 149 in Theorem 13. ◻

Proof. To estimate \(|\partial_z\nabla_{\xi}u_n-\partial_z\nabla_{\tau_1}\bar{u} |,\) firstly by straightforward calculation, we have \[\begin{align} \frac{1}{2}(\partial_z\nabla_{\xi}u_{n}^2 - \partial_z \nabla_{\tau_1} \bar{u}^2)=&\partial_zu_{n}\nabla_{\xi}u_{n}+u_{n}\partial_z\nabla_{\xi}u_{n} -\partial_z\bar{u}\nabla_{\tau_1}\bar{u}-\bar{u}\partial_z\nabla_{\tau_1}\bar{u} \\ =&\partial_zu_{n}G+u_{n}\partial_z G +\nabla_{\tau_1}\bar{u}(\partial_zu_{n}-\partial_z\bar{u}) +(u_{n}-\bar{u})\partial_z\nabla_{\tau_1}\bar{u}, \end{align}\] where \(G=\nabla_{\xi}u_{n}-\nabla_{\tau_1}\bar{u}\).

Next, we claim that \[\begin{align} |\partial_z\nabla_{\xi}u_{n}^2 - \partial_z \nabla_{\tau_1} \bar{u}^2|\leq \varepsilon^5C\phi_{1,2\alpha}u_n+ e^{-\frac{A}{x+1}} C\bar{u}^2 \quad \text{in}\quad \{z\leq \varepsilon^3\}\cap\Omega\cap\{y\leq Y^*\}. \end{align}\] If the above claim holds, then we have \[\begin{align} \label{ungback} \pm(\partial_zu_{n}G+u_{n}\partial_z G)=\pm\partial_z(u_{n} G) \leq \varepsilon^5C\phi_{1,\alpha}u_n+ e^{-\frac{A}{x+1}} C\bar{u}^2. \end{align}\tag{168}\] Then for \(z\) near \(0,\) by \(|G|\leq e^{-\frac{A}{x+1}} C{\epsilon_0}^2\) on \(z=0\), we have\[\begin{align} | G| \leq \varepsilon^5C\phi_{1,\alpha}u_n+ e^{-\frac{A}{x+1}} C\bar{u}^2. \end{align}\] Substituting this to 168 , for sufficiently small positive constant \(\delta_3\) such that \[\begin{align} \label{delta3} \delta_3=\varepsilon^{\frac{5}{1-\alpha}}, \end{align}\tag{169}\] we have \[\begin{align} \label{zeta-2025-05-23} |\partial_z\nabla_{\xi}u_{n} - \partial_z \nabla_{\tau_1} \bar{u}|\leq \varepsilon^5 e^{-\frac{A}{x+1}} C\bar{u}^{\alpha}\quad\text{in}\quad\{ z\in[0,\delta_3]\}\cap\Omega\cap\{0\leq y\leq Y^*\}. \end{align}\tag{170}\] Similarly, we can prove \[\begin{align} |\partial_z\nabla_{\eta}u_{n} - \partial_z \nabla_{\tau_1} \bar{u}|\leq \varepsilon^5 e^{-\frac{A}{x+1}} C\bar{u}^{\alpha}\quad\text{in}\quad\{ z\in[0,\delta_3]\}\cap\Omega\cap\{0\leq y\leq Y^*\}. \end{align}\]

Proof of the claim

By 80 , we have \[\begin{align} \label{ntau1} \nabla_{n}^2\bar{u}^2=4\nabla_{\tau_1} \bar{u},\quad \nabla_{n}^2\nabla_{\tau_1}\bar{u}^2= \nabla_{\tau_1}\nabla_{n}^2\bar{u}^2=4\nabla_{\tau_1} ^2\bar{u}. \end{align}\tag{171}\] By 78 , straightforward calculation yields \[\begin{align} \label{xz5}\begin{aligned} I:=u_{n}\nabla_{\psi}^2\nabla_{\xi}u_{n}^2 - \bar{u}\nabla_{n}^2 \nabla_{\tau_1} \bar{u}^2= &\nabla_{\xi} ^2u_{n}^2 +\nabla_{\xi}\tilde{q}\nabla_{\eta}u_{n}^2+(1+\tilde{q})\nabla_{\xi}\nabla_{\eta}u_{n}^2 - 2\nabla_{\tau_1} ^2 \bar{u}^2\\ & -\nabla_{\xi}u_{n}\,\nabla_{\psi}^2u_{n}^2+\nabla_{\tau_1}\bar{u}\nabla_{n}^2 \bar{u}^2 .\end{aligned} \end{align}\tag{172}\]

Step 1 We will prove 174 .

Note by 171 , \[\begin{align} | \nabla_{\xi}u_{n}\,\nabla_{\psi}^2u_{n}^2-\nabla_{\tau_1}\bar{u}\nabla_{n}^2 \bar{u}^2|\leq& | (\nabla_{\xi}u_{n}-\nabla_{\tau_1}\bar{u})\nabla_{n}^2 \bar{u}^2|+|\nabla_{\xi}u_{n}( \nabla_{\psi}^2u_{n}^2-\nabla_{n}^2 \bar{u}^2)| \\ \leq &C\bar{u} u_{n-1}\phi_{1,0}. \end{align}\] In addition, \[\begin{align} |\nabla_{\xi}\tilde{q}\nabla_{\eta}u_{n}^2 +\tilde{q}\nabla_{\xi}\nabla_{\eta}u_{n}^2 |=& |2u_{n}\nabla_{\xi}\tilde{q}\nabla_{\eta}u_{n} +2\tilde{q}(u_{n}\nabla_{\xi}\nabla_{\eta}u_{n} + \nabla_{\xi}u_{n}\nabla_{\eta}u_{n} )|\leq \varepsilon^2Cu_{n-1}^{1+\frac{\alpha}{2}}\phi_{1,0}, \end{align}\] and \[\begin{align} |\nabla_{\xi} ^2u_{n}^2 +\nabla_{\xi}\nabla_{\eta}u_{n}^2 -2\nabla_{\tau_1} ^2 \bar{u}^2| =& |2(\nabla_{\xi} u_{n})^2+2 u_{n}\nabla_{\xi}^2 u_{n} +2\nabla_{\xi} u_{n}\nabla_{\eta} u_{n}+2 u_{n}\nabla_{\xi}\nabla_{\eta} u_{n} \\&-4(\nabla_{\tau_1} \bar{u})^2 -4\bar{u}\nabla_{\tau_1}^2\bar{u}| \\ \leq & C\varepsilon^2 d_0^2\bar{u}^{\frac{\alpha}{2}} u_{n-1} \phi_{1,0}+ C\bar{u}u_{n-1}\phi_{1,0}. \end{align}\] Therefore, by 172 , we have \[\begin{align} |I|=|u_{n}\nabla_{\psi}^2\nabla_{\xi}u_{n}^2 - \bar{u}\nabla_{n}^2 \nabla_{\tau_1} \bar{u}^2| \leq C\varepsilon^2 u_{n-1}^{1+\frac{\alpha}{2}} \phi_{1,0}+ C\bar{u}u_{n-1}\phi_{1,0}. \end{align}\] On the other hand, \[\begin{align} \label{26-03-31-2}\begin{aligned} I=& u_{n}\nabla_{\psi}^2\nabla_{\xi}u_{n}^2 - \bar{u}\nabla_{n}^2 \nabla_{\tau_1} \bar{u}^2\\= & u_{n}\nabla_{\psi}(\nabla_{\psi}\nabla_{\xi}u_{n}^2 - \nabla_{n} \nabla_{\tau_1} \bar{u}^2)+ (u_{n}\nabla_{\psi}- \bar{u} \nabla_{n}) \nabla_{n} \nabla_{\tau_1} \bar{u}^2\\= & u_{n}\nabla_{\psi}(\nabla_{\psi}\nabla_{\xi}u_{n}^2 - \nabla_{n} \nabla_{\tau_1} \bar{u}^2)+\bar{u}(\frac{u_{n}}{u_{n-1}}-1)\nabla_{n}^2 \nabla_{\tau_1} \bar{u}^2 \\= & \frac{ u_{n}}{u_{n-1}}\partial_z(\frac{1}{u_{n-1}}\partial_z\nabla_{\xi}u_{n}^2 - \frac{1}{\bar{u}}\partial_z \nabla_{\tau_1} \bar{u}^2)+\bar{u}(\frac{u_{n}}{u_{n-1}}-1)\nabla_{n}^2 \nabla_{\tau_1} \bar{u}^2,\end{aligned} \end{align}\tag{173}\] where we have used that \[\begin{align} (u_{n}\nabla_{\psi}- \bar{u} \nabla_{n}) \nabla_{n} \nabla_{\tau_1} \bar{u}^2=& (\frac{u_{n}}{u_{n-1}}-1)\partial_z\nabla_{n} \nabla_{\tau_1} \bar{u}^2\\ =&\bar{u}(\frac{u_{n}}{u_{n-1}}-1)\nabla_{n}^2 \nabla_{\tau_1} \bar{u}^2. \end{align}\] Then by 171 and 173 , we have \[\begin{align} \label{26-03-31-1} |\partial_z(\frac{1}{u_{n-1}}\partial_z\nabla_{\xi}u_{n}^2 - \frac{1}{\bar{u}}\partial_z \nabla_{\tau_1} \bar{u}^2) | \leq C\varepsilon^2 u_{n-1}^{1+\frac{\alpha}{2}} \phi_{1,0}+ C\bar{u}u_{n-1}\phi_{1,0}. \end{align}\tag{174}\]

Step 2 We will prove \(|\frac{1}{u_{n-1}}\partial_z\nabla_{\xi}u_{n}^2 - \frac{1}{\bar{u}}\partial_z \nabla_{\tau_1} \bar{u}^2|\leq \varepsilon^2 e^{-\frac{A}{x+1}}C{\epsilon_0}\) at \(z=0\), through the compatible boundary condition.

By the definition of \(\nabla_{\psi}\) and 32 , we have \[\begin{align} \label{26-03-30-7}\begin{aligned} \frac{1}{u_{n-1}}\partial_z\nabla_{\xi}u_{n}^2 - \frac{1}{\bar{u}}\partial_z \nabla_{\tau_1} \bar{u}^2=&\nabla_{\psi}\nabla_{\xi}u_{n}^2 - \nabla_{n} \nabla_{\tau_1} \bar{u}^2\\= & \nabla_{\xi}\nabla_{\psi}u_{n}^2 - \nabla_{\tau_1} \nabla_{n} \bar{u}^2\\ = & \nabla_{\xi}(2\frac{u_n}{u_{n-1}}\partial_zu_{n}) - \nabla_{\tau_1} (2\partial_z\bar{u}).\end{aligned} \end{align}\tag{175}\] And by the definition of \(\nabla_{\xi}\), we have when \(z=0,\) \[\begin{align} \label{xz6}\begin{aligned} \nabla_{\xi}(\frac{u_n}{u_{n-1}}\partial_zu_{n})=&\partial_x(\frac{u_n}{u_{n-1}}\partial_zu_{n}) -\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\partial_z(\frac{u_n}{u_{n-1}}\partial_zu_{n}) \\=&\partial_x(\frac{u_n}{u_{n-1}}\partial_zu_{n})\end{aligned} \end{align}\tag{176}\] because \(\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}=0\) when \(z=0.\) So does \(\nabla_{\tau_1}\partial_z \bar{u}=\partial_{x}\partial_z \bar{u}\) at \(z=0.\) Substituting back to 175 , we complete step 2 by the compatible boundary condition.

Then combining step 1 and step 2, by multiplying \(\bar{u}\) to the integral of 174 , we have \[\begin{align} |\frac{\bar{u}}{u_{n-1}}\partial_z\nabla_{\xi}u_{n}^2 - \partial_z \nabla_{\tau_1} \bar{u}^2|\leq C\bar{u}^2(\varepsilon^2 u_{n-1}^{1+\frac{\alpha}{2}} \phi_{1,0}+ C\bar{u}u_{n-1}\phi_{1,0})+ e^{-\frac{A}{x+1}} C\bar{u}^2. \end{align}\] Since \(|\frac{\bar{u}}{u_{n-1}}-1|\leq \varepsilon^5\phi_{1,2\alpha}\) and \[\begin{align} |\partial_z\nabla_{\xi}u_{n}^2|=&|2\partial_z(u_{n}\nabla_{\xi}u_{n})| =|2\partial_zu_{n}\nabla_{\xi}u_{n}+2u_{n}\partial_z\nabla_{\xi}u_{n}| \leq Cu_ne^{-\frac{(z+{\epsilon_0})^2}{x+1}\mu}, \end{align}\] we have \[\begin{align} |\partial_z\nabla_{\xi}u_{n}^2 - \partial_z \nabla_{\tau_1} \bar{u}^2|\leq \varepsilon^5C\phi_{1,2\alpha}u_n+ e^{-\frac{A}{x+1}}C\bar{u}^2\quad \text{in}\quad \{z\leq \varepsilon^3\}\cap\Omega\cap\{y\leq Y^*\} \end{align}\] which gives the claim. And then we complete the proof of theorem. ◻

Proof. Set \[H=\nabla_{\eta,\xi}u_n-\nabla_{\tau_1}\bar{u}.\] We estimate \(\nabla_{\psi }H\) in \((0,X]\times(0,Y^*]\times[\delta_3,+\infty).\) By 101 , \[\begin{align} \label{h0} P_1H-\nabla_{\psi}H\nabla_{\psi}u_{n}-H\nabla_{\psi}^2u_{n}=R_2. \end{align}\tag{177}\] Observe that by 77 and the bootstrap assumption, it holds \[\begin{align} \label{h1}\begin{aligned}&|\nabla_{\psi}^2u_{n}|+|\nabla_{\psi}^3u_{n}| \leq C_{\delta_3}e^{-\frac{(z+{\epsilon_0})^2}{x+1}\mu}, \\&|\nabla_{\psi }\nabla_{\xi }H|+|\nabla_{\psi }\nabla_{\eta }H|+|\nabla_{\psi }H|+|\nabla_{\xi }H|+|\nabla_{\eta }H|+|H| \leq C_{\delta_3}\phi_{1,0}, \\& \text{in} \quad [0,X]\times[0,Y^*]\times[\delta_3,+\infty).\end{aligned} \end{align}\tag{178}\] Then by 177 , we have \[\begin{align} \label{h2} |\nabla_{\psi }^2H|\leq& C_{\delta_3}\phi_{1,0}\quad \text{in} \quad [0,X]\times[0,Y^*]\times[\delta_3,+\infty). \end{align}\tag{179}\] By 152 , we have \[\begin{align} \label{h21} |\nabla_{\psi }^2\nabla_{\eta}\tilde{q}|\leq& C_{\delta_3}\phi_{1,0}\quad \text{in} \quad [0,X]\times[0,Y^*]\times[\delta_3,+\infty). \end{align}\tag{180}\] Combining 178 180 and 42 , we have \[\begin{align} \label{h3} |\nabla_{\psi }R_2|\leq& C_{\delta_3}\phi_{1,0}\quad \text{in} \quad [0,X]\times[0,Y^*]\times[\delta_3,+\infty). \end{align}\tag{181}\] Therefore, by 177 , \[\begin{align} \label{h4} P_2\nabla_{\psi}H=R_3, \end{align}\tag{182}\] with \[\begin{align} \label{h5} |R_3|\leq& C_{\delta_3}\phi_{1,0}\quad \text{in} \quad [0,X]\times[0,Y^*]\times[\delta_3,+\infty), \end{align}\tag{183}\] where we have used \[\begin{align} \nabla_{\psi}\nabla_{\eta}=\nabla_{\eta} \nabla_{\psi}- \frac{ \nabla_{\eta}\tilde{q}}{1+\tilde{q}}\nabla_{\psi}, \end{align}\] which is derived by \((1+\tilde{q})\nabla_{\tilde{\psi}}=\nabla_{\psi}\) in 32 . Moreover, by 76 , \(P_2 u_n=( \nabla_{\psi}u_n)^2.\) Then \[\begin{align} \label{565} P_2(u_n\nabla_{\psi}H)=R_4, \end{align}\tag{184}\] with \[\begin{align} \label{h10}\begin{aligned} R_4:=&u_nR_3+( \nabla_{\psi}u_n)^2\nabla_{\psi}H-2u_n\nabla_{\psi}u_n\nabla_{\psi}^2H,\\ |R_4|\leq& C_{\delta_3}\phi_{1,0}\quad \text{in} \quad [0,X]\times[0,Y^*]\times[\delta_3,+\infty),\end{aligned} \end{align}\tag{185}\] where we have used 178 179 and 183 . Set \[g=\pm u_n\nabla_{\psi}H-\varepsilon^{\frac{9}{2}}\phi_{1,0}.\] Then by 170 , we have \(g<0\) in \(z\in[0,\delta_3].\) Take \(A\gg C_{\delta_{3}}.\) Then \(P_2 g<0\) in \((0,X]\times(0,Y^*]\times[\delta_3,+\infty)\). Then we have ?? by the boundary condition and the maximum principle. ◻

Proof. Set \[\hat{H}=(\nabla_{\eta,\xi}u_n-\nabla_{\eta,\xi}\bar{u})\varphi_{\delta_\varepsilon},\] where \(\varphi_{\delta_\varepsilon}(z)\) is a smooth cutoff function such that \(0\leq \varphi_{\delta_\varepsilon}\leq 1\), \(\varphi_{\delta_\varepsilon}=1\) for \(z\geq \delta_\varepsilon\) and \(\varphi_{\delta_\varepsilon}=0\) for \(0\leq z\leq \frac{\delta_\varepsilon}{2},\) \(|\partial^j_z\varphi_{\delta_\varepsilon}|\leq C_{\delta_\varepsilon},\) \(j=1,2,3\). Here, \(C_{\delta_\varepsilon}\) is a positive constant depending on \(\delta_\varepsilon.\) Then by 184 , \[\begin{align} P_2(u_n\nabla_{\psi}\hat{H})=&R_5, \end{align}\] where \[\begin{align} R_5=&R_4\varphi_{\delta_\varepsilon}+P_2(\varphi_{\delta_\varepsilon}) u_n\nabla_{\psi}H-2u_n\nabla_{\psi}\varphi_{\delta_\varepsilon}\nabla_{\psi}(u_n\nabla_{\psi}H) \\& + P_2\big(\frac{u_n}{u_{n-1}}\partial_z(\nabla_{\tau_1}\bar{u}-\nabla_{\eta,\xi}\bar{u})\varphi_{\delta_\varepsilon}\big) \\&+P_2(u_n(\nabla_{\eta,\xi}u_n-\nabla_{\eta,\xi}\bar{u})\nabla_\psi\varphi_{\delta_\varepsilon}). \end{align}\] Then \[\begin{align} |R_5|\leq& C_{\delta_\varepsilon}\phi_{1,0}\quad \text{in} \quad [0,X]\times[0,Y^*]\times[\delta_\varepsilon,+\infty). \end{align}\] Set \[g=\pm u_n\nabla_{\psi}\hat{H}-\varepsilon^{\frac{9}{2}}\phi_{1,0}.\] By taking \(A\gg C_{\delta_{\varepsilon}},\) we have \(P_2 g<0\) in \((0,X]\times(0,Y^*]\times[\delta_\varepsilon,+\infty)\). Then by the boundary data and the maximum principle, we complete the proof of the theorem. ◻

Proof. By 22 , we have \[\begin{align} \label{68}\begin{aligned} \partial_{z}u_n^2= &[\int_0^z2u_{n-1}(\nabla_{\xi} u_n +(1+\tilde{q})\nabla_{\eta} u_n )dz']u_{n-1}+(\frac{1}{u_{n-1}}\partial_{z}u_n^2)|_{z=0}u_{n-1} \quad \text{in} \quad \Omega. \end{aligned} \end{align}\tag{186}\] Then \[\begin{align} \begin{aligned} \partial_{z}u_n= &[\int_0^zu_{n-1}(\nabla_{\xi} u_n +(1+\tilde{q})\nabla_{\eta} u_n )dz']\frac{u_{n-1}}{u_n}+(\frac{u_n}{u_{n-1}}\partial_{z}u_n)|_{z=0}\frac{u_{n-1}}{u_n} \quad \text{in} \quad \Omega.\end{aligned} \end{align}\] This implies \[\begin{align} \label{xzx1}\begin{aligned} \partial_{z}^2u_n= &\partial_{z}\{[\int_0^zu_{n-1}(\nabla_{\xi} u_n +(1+\tilde{q})\nabla_{\eta} u_n )dz']\frac{u_{n-1}}{u_n}\}\\&+(\frac{u_n}{u_{n-1}}\partial_{z}u_n)|_{z=0}\partial_{z}(\frac{u_{n-1}}{u_n} ),\\ \partial_{z}^2\bar{u}= &2\bar{u}\nabla_{\tau_1} \bar{u}. \end{aligned} \end{align}\tag{187}\] Therefore, \[\begin{align} \label{1116-3-exp}\begin{aligned} \partial_{z}^2(u_n-\bar{u})= &\partial_{z}\{[\int_0^zu_{n-1}(\nabla_{\xi} u_n +(1+\tilde{q})\nabla_{\eta} u_n )dz']\frac{u_{n-1}}{u_n}\}-2\bar{u}\nabla_{\tau_1} \bar{u} \\&+(\frac{u_n}{u_{n-1}}\partial_{z}u_n)|_{z=0}\partial_{z}(\frac{u_{n-1}}{u_n} ). \end{aligned} \end{align}\tag{188}\] We claim that the first term for \(\partial_{z}^2(u_n-\bar{u})\) in 188 satisfies \[\begin{align} \label{1029} |\partial_{z}\{[\int_0^zu_{n-1}(\nabla_{\xi} u_n +(1+\tilde{q})\nabla_{\eta} u_n )dz']\frac{u_{n-1}}{u_n}\}-2\bar{u}\nabla_{\tau_1} \bar{u}|\leq \varepsilon^2 C\phi_{1,1}. \end{align}\tag{189}\]

We prove the claim 189 as follows. By straightforward calculation, we have \[\begin{align} &\partial_{z}[\int_0^zu_{n-1}\nabla_{\xi} u_ndz'\frac{u_{n-1}}{u_n}]-\bar{u}\nabla_{\tau_1} \bar{u} \\ =&u_{n-1}\nabla_{\xi} u_n\frac{u_{n-1}}{u_n}-\bar{u}\nabla_{\tau_1} \bar{u}+\int_0^zu_{n-1}\nabla_{\xi} u_ndz' (\frac{\partial_z(u_{n-1}-u_n)}{u_n}-\frac{\partial_zu_n(u_{n-1}-u_n)}{u_n^2}) \\ =&u_{n-1}\nabla_{\xi} u_n\frac{u_{n-1}-u_n}{u_n}+u_{n-1}\nabla_{\xi} u_n-\bar{u}\nabla_{\tau_1} \bar{u}+\int_0^zu_{n-1}\nabla_{\xi} u_ndz' (\frac{\partial_z(u_{n-1}-u_n)}{u_n}-\frac{\partial_zu_n(u_{n-1}-u_n)}{u_n^2}) \\ =&u_{n-1}\nabla_{\xi} u_n\frac{u_{n-1}-u_n}{u_n}+(u_{n-1}-\bar{u})\nabla_{\xi} u_n+\bar{u}(\nabla_{\xi} u_n-\nabla_{\tau_1} \bar{u}) \\&+\int_0^zu_{n-1}\nabla_{\xi} u_ndz' (\frac{\partial_z(u_{n-1}-u_n)}{u_n}-\frac{\partial_zu_n(u_{n-1}-u_n)}{u_n^2}). \end{align}\] Since \(|\int_0^zu_{n-1}\nabla_{\xi} u_ndz'|\leq C\bar{u}^2,\) by ?? , ?? and induction assumption, we have \[\begin{align} \label{1117-1} |\partial_{z}[\int_0^zu_{n-1}\nabla_{\xi} u_ndz'\frac{u_{n-1}}{u_n}]-\bar{u}\nabla_{\tau_1} \bar{u}|\leq \varepsilon^2 C\phi_{1,1}\quad\text{in}\quad \Omega\cap\{0\leq y\leq Y^*\} . \end{align}\tag{190}\] We can estimate the other terms in 189 similarly by using \(|\tilde{q}|\leq\varepsilon^2 C\phi_{1,\alpha}\) and the induction assumption. Then the claim 189 holds.

Next, we estimate the other terms for \(\partial_{z}^2(u_n-\bar{u})\) in 188 . By Lemma 20 in Appendix, we have, for a small positive constant \(\delta_5\) independent of \(\varepsilon,\) \[\begin{align} \label{1117-2} \big|(\frac{u_n}{u_{n-1}}\partial_{z}u_n)|_{z=0}\partial_{z}(\frac{u_{n-1}}{u_n} )\big|\leq \frac{\varepsilon e^{-\frac{A}{x+1}}(z+\frac{u_n(x,y,0)}{\partial_zu_n(x,y,0)})^{\alpha}}{1+\frac{\alpha}{5}}\quad \text{in}\quad \Omega\cap\{0\leq y\leq Y^*\}\cap\{z\leq \delta_5\}. \end{align}\tag{191}\] Combining 190 and 191 , we have, \[\begin{align} |\partial_{z}^2(u_n-\bar{u})|\leq \frac{\varepsilon e^{-\frac{A}{x+1}}}{1+\frac{\alpha}{6}}\min\{1,(z+ \frac{\bar{u}(x,y,0)}{\partial_z\bar{u}(x,y,0)})^\alpha\} \quad \text{in} \quad[0,X]\times[0,Y^*]\times[0,\delta_5]. \end{align}\] On the other hand, by \(\partial_z \bar{u}>0\), 44 and 56 , we have \(u_{n-1}, u_n\geq c_0\delta_5\) in \([0,X]\times[0,Y^*]\times[\delta_5,+\infty).\) Then by ?? and ?? , \[\begin{align} |\partial_z (\frac{u_{n-1}}{u_n})|=|\partial_z (\frac{u_n-u_{n-1}}{u_n})|\leq \varepsilon^3 \phi_{1,\alpha} \quad \text{in} \quad [0,X]\times[0,Y^*]\times[\delta_5,+\infty). \end{align}\] Then \(|\partial_{z}^2(u_n-\bar{u})|\leq \frac{\varepsilon e^{-\frac{A}{x+1}}}{1+\frac{\alpha}{6}}(z+ \frac{\bar{u}(x,y,0)}{\partial_z\bar{u}(x,y,0)})^\alpha\) in \([0,X]\times[0,Y^*]\times[\delta_5,+\infty)\) because \[|(\frac{u_n}{u_{n-1}}\partial_{z}u_n)|_{z=0}|\leq C\quad \text{in} \quad[0,X]\times[0,Y^*].\] In summary, \(|\partial_{z}^2(u_n-\bar{u})|\leq \frac{\varepsilon e^{-\frac{A}{x+1}}}{1+\frac{\alpha}{6}}\min\{1,(z+ \frac{\bar{u}(x,y,0)}{\partial_z\bar{u}(x,y,0)})^\alpha\}\) in \([0,X]\times[0,Y^*]\times[0,+\infty).\) And this completes the proof of the theorem. ◻

Proof. By Theorem 5 and Theorem 7, we can improve the estimates for \(u_{n-1}-\bar{u}, v_{n-1}-\bar{u}, \partial_z(u_{n-1}-\bar{u})\) as follows. \[\begin{align} | u_{n-1}-\bar{u} |+| v_{n-1}-\bar{u} |\leq Cd_0\varepsilon^6 \phi_{1,0},\quad |\partial_z u_{n-1}-\partial_z\bar{u}|\leq \frac{d_0\varepsilon^4}{4} \phi_{1,\alpha}\quad \text{in} \quad \Omega. \end{align}\] Then for a small positive constant \(\alpha,\) \[\begin{align} \label{ubdep}\begin{aligned} |\frac{u_{n-1}}{u_{n}} -1|=&\frac{|u_{n-1}-u_{n}|}{u_{n}}\leq C\varepsilon^{4.5} \phi_{1,0} ,\quad\frac{|u_{n-1}-u_{n}|}{u_{n-1}^2}\leq C\varepsilon^3 \phi_{1,0},\\ \frac{ |\partial_z u_n-\partial_zu_{n-1}|}{u_{n-1}}\leq & C\varepsilon^{2.5} \phi_{1,\alpha}, \quad |\tilde{q}|=|\frac{q_{n-1}}{u_{n-1}}|\leq C\varepsilon^4 \phi_{1,0}\quad\text{in}\quad \Omega\cap \{z\geq (\varepsilon d_0)^{\frac{1}{1-\alpha}}\},\end{aligned} \end{align}\tag{192}\] because \(d_0\) is independent of \(\varepsilon\) and \[\begin{align} \label{lbdep} u_{n-1}, u_{n}, \bar{u}\geq c_0(\varepsilon d_0)^{\frac{1}{1-\alpha}}\quad\text{in}\quad \Omega\cap \{z\geq (\varepsilon d_0)^{\frac{1}{1-\alpha}}\}. \end{align}\tag{193}\] Recall \(c_0\) in 47 . Then \[\begin{align} \label{1104}\begin{aligned} |\partial_z (\frac{ u_n}{u_{n-1}})|=&|\partial_z (\frac{ u_n-u_{n-1}}{u_{n-1}})|\\ \leq &\frac{ |\partial_z u_n-\partial_zu_{n-1}|}{u_{n-1}}+\frac{\partial_zu_{n-1}|u_{n-1}-u_{n}|}{u_{n-1}^2} \\ \leq& C\varepsilon^{2.5} \phi_{1,0}\quad\text{in}\quad \Omega\cap \{z\geq (\varepsilon d_0)^{\frac{1}{1-\alpha}}\}.\end{aligned} \end{align}\tag{194}\] Next, by 22 , we have \[\begin{align} \begin{aligned} \partial_{z}^2u_n= &[u_{n-1}(\nabla_{\xi} u_n +(1+\tilde{q})\nabla_{\eta} u_n )-\partial_{z}u_n \partial_{z}(\frac{u_n}{u_{n-1}} )]\frac{u_{n-1}}{u_{n}},\\ \partial_{z}^2\bar{u}= &2\bar{u}\nabla_{\tau_1} \bar{u} . \end{aligned} \end{align}\] Then by Theorem 10, i.e. \(|\nabla_{\eta,\xi}u_n-\nabla_{\tau_1}\bar{u} |\leq C\varepsilon^5 \phi_{1,0}\) in \(\Omega,\) 192 and 194 , we have \[\begin{align} | \partial_{z}^2(u_n-\bar{u})|\leq C\varepsilon^{2.5} \phi_{1,0}\quad\text{in}\quad \Omega\cap \{z\geq (\varepsilon d_0)^{\frac{1}{1-\alpha}}\}. \end{align}\] In particular, for \(\varepsilon\) small enough, we have \[\begin{align} C\varepsilon^{0.2} \phi_{1,0} \leq\phi_{1,\alpha},\quad (\varepsilon d_0)^{\frac{1}{1-\alpha}}\leq z\leq \delta, \end{align}\] because for \(\frac{\alpha}{1-\alpha}\leq 0.1,\) it holds \[C\varepsilon^{0.2}\leq \frac{1}{2}(\frac{1}{\sqrt{X+1}})^\alpha(\varepsilon d_0)^{\frac{\alpha}{1-\alpha}} \leq (\frac{z+\epsilon_0}{\sqrt{x+1}})^\alpha,\quad (\varepsilon d_0)^{\frac{1}{1-\alpha}}\leq z\leq \delta,\] where \(\delta\) is defined in 45 . And this completes the proof of the lemma. ◻

Proof of Lemma 9. Based on the transformation formula in 35 , we have \[\begin{align} \label{1105-1} \partial_{z}\partial_{x}=& \partial_{z}\nabla _{\xi} +\partial_{z}(\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}})\partial_z +\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\partial_z^2. \end{align}\tag{210}\] By induction assumption, we have \[\begin{align} \label{dz1} |\partial_z(\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}})|\leq C, \quad |\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}|\leq C\min\{z,1\}. \end{align}\tag{211}\] Then by 56 and 46 , we have \[\begin{align} |\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\partial_z^2(u_n-\bar{u})|\leq& Cz\varepsilon e^{-\frac{A}{x+1}}(z+\epsilon_0)^{\alpha}\\ \leq& C\varepsilon^4 e^{-\frac{A}{x+1}}(z+\epsilon_0)^{\alpha} \quad \text{in} \quad \Omega\cap\{0\leq z\leq \varepsilon^3\}. \end{align}\]

We now consider the estimate in two regions.

Step 1 Estimate in \(\Omega\cap\{0\leq z\leq \delta_\varepsilon\}.\) Here, \(\delta_{\varepsilon}\) is defined in Remark 15 with \(\delta_{\varepsilon}\leq \varepsilon^3.\) Firstly, by the above estimates, \[\begin{align} |\partial_{z}(\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}})\partial_z(u_n-\bar{u}) +\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\partial_z^2(u_n-\bar{u})|\leq \varepsilon^{3}\phi_{1,\alpha} \quad \text{in} \quad \Omega\cap\{0\leq z\leq \delta_\varepsilon\}. \end{align}\] Next, by ?? , we have \[\begin{align} |\partial_z\nabla_{\eta,\xi}u_n-\partial_z\nabla_{\eta,\xi}\bar{u} |\leq \frac{ \varepsilon^2}{2+\frac{\alpha}{3}} \phi_{1,\alpha} \quad \text{in} \quad \Omega\cap\{0\leq z\leq \delta_\varepsilon\}. \end{align}\] By 210 , we have \[\begin{align} \label{b4} |\partial_z\partial_x(u_n-\bar{u}) |\leq \frac{ \varepsilon^2}{2} \phi_{1,\alpha} \quad \text{in} \quad \Omega\cap\{0\leq z\leq \delta_\varepsilon\}. \end{align}\tag{212}\]

Step 2 Estimate in \(\Omega\cap\{z\geq \delta_\varepsilon\}.\) In this region, by 211 , we have \[\begin{align} |\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}|\leq Cd_0\varepsilon\quad \text{in }\quad \Omega\cap \{0\leq z\leq\varepsilon d_0\}. \end{align}\]Then by 56 , we have for small \(d_0\) independent of \(\varepsilon,\) \[\begin{align} \label{b1-17} |\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\partial_z^2(u_n-\bar{u}) | \leq Cd_0\varepsilon^2\phi_{1,\alpha}\leq \frac{\varepsilon^2}{3}\phi_{1,\alpha}\quad \text{in }\quad \Omega\cap \{0\leq z\leq\varepsilon d_0\}. \end{align}\tag{213}\] By Lemma 8, we have \[\begin{align} \label{26-04-02-01} | \partial_{z}^2(u_n-\bar{u})|\leq C\varepsilon^{2.3} \phi_{1,\alpha}\quad\text{in}\quad \Omega\cap \{z\geq \varepsilon d_0\}, \end{align}\tag{214}\] because \(d_0\varepsilon<1\) so that \(\varepsilon d_0>(\varepsilon d_0)^{\frac{1}{1-\alpha}}.\) Then by 211 , 213 and 214 , we obtain \[\begin{align} \label{b1} |\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\partial_z^2(u_n-\bar{u}) | \leq Cd_0\varepsilon^2\phi_{1,\alpha}\leq \frac{\varepsilon^2}{3}\phi_{1,\alpha}\quad \text{in }\quad \Omega\cap \{z\geq \delta_\varepsilon\}. \end{align}\tag{215}\] Moreover, by Lemma 7 and 56 , we have \[\begin{align} \label{b2} |\partial_z\nabla_{\eta,\xi}u_n-\partial_z\nabla_{\eta,\xi}\bar{u} |\leq \varepsilon^4 \phi_{1,\alpha} \quad \text{in} \quad \Omega\cap\{z\geq\delta_\varepsilon\}, \end{align}\tag{216}\] and \[\begin{align} \label{b3} |\partial_{z}(\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}})\partial_z(u_n-\bar{u})|\leq \varepsilon^3\phi_{1,\alpha}\quad \text{in} \quad \Omega. \end{align}\tag{217}\] Finally, by 210 , we have \[\begin{align} \label{b5} |\partial_z\partial_x(u_n-\bar{u}) |\leq \frac{ \varepsilon^2}{2} \phi_{1,\alpha} \quad \text{in} \quad \Omega\cap\{ z\geq \delta_\varepsilon\}. \end{align}\tag{218}\] Combining 212 and 218 completes the proof of the lemma. ◻

Proof. By 103 , we have \[\begin{align} \begin{aligned} ( P_2+ P_3)(\nabla_{\xi}u_n-\nabla_{\xi}\bar{u} ) =R , \end{aligned} \end{align}\] with \[\begin{align} R=&R_2+( P_2+ P_3)( \nabla_{\tau_1}\bar{u}-\nabla_{\xi}\bar{u} ),\\ |R|\leq &C_{\delta}\phi_{1,0} \quad \text{in} \quad \Omega\cap \{z\geq \delta\}, \end{align}\] by 227 where \(C_\delta\) is a positive constant depending on \(\delta\). Since \(u_n, u_{n-1}, \bar{u}\geq c_0 \delta\) in \(\Omega\cap \{z\geq \delta\}\) and the properties of the background profile \(\bar{u}\) are known, by taking \(A\) large depending on \(C_{\delta}\) and applying the maximum principle as in the proof of Theorem 10, we have the estimate stated in the lemma. ◻

Proof of Lemma 10. By Remark 19, we have \[\begin{align} |\partial_x(u_n-\bar{u}) |\leq \frac{\varepsilon^2}{2(1+\alpha)}\sqrt{x+1} \phi_{1,\alpha+1}+\epsilon_0^2<\varepsilon^2\phi_{1,1},\quad z+\epsilon_0\leq \delta, \end{align}\] because \(x\leq X\leq \theta\) with \(\theta\) small.

For \(\varepsilon\) small, we have \[\begin{align} \label{yw5} \varepsilon\phi_{1,0}\leq \phi_{1,1},\quad z+\epsilon_0\geq \delta. \end{align}\tag{219}\] Then by \[\begin{align} \label{1105-2} \partial_{x}=& \nabla _{\xi} +\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\partial_z \end{align}\tag{220}\] from 35 , Lemma 12, 219 , 211 and 56 , we have \[\begin{align} |\partial_x(u_n-\bar{u}) |\leq \varepsilon^2 \phi_{1,1},\quad z+\epsilon_0\geq \delta. \end{align}\] ◻

Proof of Lemma 11. By 35 , we have \[\begin{align} \label{1105-3} \begin{aligned} \partial_{x}^2=& (\nabla _{\xi} +\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\partial_z)\partial_{x}\\ =& \nabla _{\xi} \partial_{x}+\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\partial_z\partial_{x}\\ =& \nabla _{\xi} (\nabla _{\xi} +\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\partial_z)+\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\partial_z\partial_{x}\\ =& \nabla _{\xi}^2 +\nabla _{\xi}(\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}})\partial_z+\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\nabla _{\xi}\partial_z+\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\partial_z\partial_{x}\\ =& \nabla _{\xi}^2 +\nabla _{\xi}(\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}})\partial_z+\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\partial_z\partial_{x} \\& +\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}[(\frac{\nabla _{\xi}u_{n-1}}{u_{n-1}})\partial_z+\partial_z\nabla _{\xi}], \end{aligned} \end{align}\tag{221}\] where we have used \[\begin{align} \nabla _{\xi}\partial_z=\nabla _{\xi}(u_{n-1}\nabla _{\psi}) =\nabla _{\xi}u_{n-1}\nabla _{\psi}+u_{n-1}\nabla _{\psi}\nabla _{\xi} =(\frac{\nabla _{\xi}u_{n-1}}{u_{n-1}})\partial_z+\partial_z\nabla _{\xi}. \end{align}\] By the induction assumption, it holds \[|\nabla _{\xi}(\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}})|\leq C\min\{(z+\epsilon_0)^{\frac{\alpha}{2}},1\},\quad |\frac{\nabla _{\xi}u_{n-1}}{u_{n-1}}|\leq C,\] and \[\begin{align} |\frac{\int_0^z \partial_{x}u_{n-1}dz'\nabla _{\xi}u_{n-1}}{u_{n-1}^2}|\leq C\min\{z,1\}. \end{align}\] Then by Corollary 1 with \(d_0\) in ?? being small and 56 , we have \[\begin{align} \label{117}\begin{aligned} &|\nabla _{\xi}^2(u_n-\bar{u}) +\nabla _{\xi}(\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}})\partial_z(u_n-\bar{u}) +\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}(\frac{\nabla _{\xi}u_{n-1}}{u_{n-1}})\partial_z(u_n-\bar{u})| \\ \leq&\varepsilon^2d_0\phi_{1,\frac{\alpha}{2}}\quad \text{in }\quad \Omega. \end{aligned} \end{align}\tag{222}\] By 211 , ?? , Lemma 7 and ?? , we have \[\begin{align} |\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\big(\partial_z\partial_{x}(u_n-\bar{u}) +\partial_z\nabla _{\xi}(u_n-\bar{u})\big)|\leq d_0C\varepsilon^2 \phi_{1,\alpha}\quad \text{in }\quad \Omega\cap\{z\leq d_0\}. \end{align}\] On the other hand, by 210 , 211 , 56 and Lemma 8,\[\begin{align} \begin{aligned} &|\partial_{z}\partial_{x}(u_n-\bar{u})- \partial_{z}\nabla _{\xi} (u_n-\bar{u})| \\ =& |\partial_{z}\big(\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\partial_{z}(u_n-\bar{u})\big)|\\ =& |\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\partial_{z}^2(u_n-\bar{u})|+ |\partial_{z}(\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}})\partial_{z}(u_n-\bar{u})|\\ \leq&\varepsilon^{2461}\phi_{1,\alpha} \quad\Omega\cap\{z\geq d_0\}. \end{aligned} \end{align}\] Then by Lemma 7, \[\begin{align} |\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\big(\partial_z\partial_{x}(u_n-\bar{u}) +\partial_z\nabla _{\xi}(u_n-\bar{u})\big)|\leq C\varepsilon^{2461} \phi_{1,\alpha}\quad \text{in }\quad \Omega\cap\{z\geq d_0\}. \end{align}\] In summary, we have \[\begin{align} |\frac{\int_0^z \partial_{x}u_{n-1}dz'}{u_{n-1}}\big(\partial_z\partial_{x}(u_n-\bar{u}) +\partial_z\nabla _{\xi}(u_n-\bar{u})\big)|\leq d_0C\varepsilon^2 \phi_{1,\alpha}\quad \text{in }\quad \Omega. \end{align}\] Combining this with 222 and 221 , we have \[\begin{align} \begin{aligned} |\partial_{x}^2u_{n}-\partial_{x}^2\bar{u}|\leq C d_0\varepsilon^2 \phi_{1,\frac{\alpha}{2}}<\varepsilon^2 \phi_{1,\frac{\alpha}{2}}\quad \text{in }\quad\Omega.\end{aligned} \end{align}\] ◻

Proof. By 223 , for small \(\varepsilon,\) \(\alpha\in(0,1)\) and some small positive constants \(\lambda\) and \(\delta\), we have \[\begin{align} \begin{aligned} P_1 z^{\alpha}=& -\tilde{b}\alpha z^{\alpha-1} +\alpha(-\frac{\partial_zu_{n}}{u_{n-1}^2} +\frac{u_{n}\partial_zu_{n-1}}{u_{n-1}^3})z^{\alpha-1} -\frac{u_n}{u_{n-1}^2}\alpha(\alpha-1)z^{\alpha-2}\\ \geq& -C\alpha z^\alpha +\frac{1}{2}\frac{u_n}{u_{n-1}^2}\alpha(-\alpha+1)z^{\alpha-2} \\ \geq &\frac{\lambda}{u_{n-1}}\alpha(-\alpha+1)z^{\alpha-2}, \quad z\leq \delta, \end{aligned} \end{align}\] where we have used \[\begin{align} |\frac{u_n}{u_{n-1}^2}-\frac{1}{u_{n-1}}|= |\frac{u_n-u_{n-1}}{u_{n-1}^2}|\leq \frac{\varepsilon e^{-\frac{A}{x+1}} C}{u_{n-1}}, \end{align}\] and \[\begin{align} | -\frac{\partial_zu_{n}}{u_{n-1}^2} +\frac{u_{n}\partial_zu_{n-1}}{u_{n-1}^3}|=&| \frac{-\partial_zu_{n}u_{n-1}+u_{n}\partial_zu_{n-1}}{u_{n-1}^3}|\\ =&| \frac{(\partial_zu_{n-1}-\partial_zu_{n})u_{n-1} +(-u_{n-1} +u_{n})\partial_zu_{n-1}}{u_{n-1}^3}|\\ \leq & \frac{\varepsilon e^{-\frac{A}{x+1}}C}{u_{n-1}^2}, \end{align}\]by 223 . ◻

Proof. Since \(u_{n-1}\psi_{n-1}^{\alpha-2}\geq \lambda_0 \psi_{n-1}^{\alpha-\frac{3}{2}}\), \(z\leq \delta_0\) for some small positive constants \(\lambda_0\) and \(\delta_0\), by 223 , 59 and 60 , for \(\alpha\in(0,1),\) it holds \[\begin{align} \begin{aligned} P_2\psi_{n-1}^{\alpha}= &\nabla_\xi \psi_{n-1}^\alpha +(1+\tilde{q})\nabla_{\eta} \psi_{n-1}^\alpha-u_{n}\nabla_{\psi}^2\psi_{n-1}^\alpha \\ =&\alpha(\tilde{q}\int_0^z \partial_yu_{n-1}dz'- \int_0^z \partial_yq_{n-1}dz')\psi_{n-1}^{\alpha-1}+\alpha(-\alpha+1)u_{n}\psi_{n-1}^{\alpha-2} \\ \geq &\lambda\psi_{n-1}^{\alpha-\frac{3}{2}} ,\\ P_2(\psi_{n-1}-\psi_{n-1}^{1+\alpha})=&-\nabla_\xi \psi_{n-1}^{1+\alpha} +(1+\tilde{q})\nabla_{\eta} (\psi_{n-1}-\psi_{n-1}^{1+\alpha})+u_{n}\nabla_{\psi}^2\psi_{n-1}^{1+\alpha} \\ =&(\tilde{q}\int_0^z \partial_yu_{n-1}dz'- \int_0^z \partial_yq_{n-1}dz')(1-(1+\alpha)\psi_{n-1}^{\alpha})\\&+\alpha(\alpha+1)u_{n}\psi_{n-1}^{\alpha-1} \\ \geq &\lambda\psi_{n-1}^{\alpha-\frac{1}{2}} , \end{aligned} \end{align}\] for \(z\leq \delta\) such that \(\psi_{n-1}\ll1.\) ◻

Proof. By 70 and 223 , for \(\mu\) small enough and for \(z\) large enough, it holds \[\begin{align} P_1e^{-\frac{z^2}{x+1}\mu}=&\partial_x e^{-\frac{z^2}{x+1}\mu} -\tilde{b}\partial_{z} e^{-\frac{z^2}{x+1}\mu} -\frac{u_n}{u_{n-1}^2}\partial_{z}^2e^{-\frac{z^2}{x+1}\mu} +(-\frac{\partial_zu_{n}}{u_{n-1}^2} +\frac{u_{n}\partial_zu_{n-1}}{u_{n-1}^3}) (-\frac{2z}{x+1}\mu)e^{-\frac{z^2}{x+1}\mu} \\ \geq &e^{-\frac{z^2}{x+1}\mu} [\frac{z^2}{(x+1)^2}\mu -C\frac{2z}{x+1}\mu -\frac{u_n}{u_{n-1}^2}(\frac{4z^2}{(x+1)^2}\mu^2-\frac{2}{x+1}\mu) ]\\ \geq& \frac{1}{2}e^{-\frac{z^2}{x+1}\mu} \frac{z^2}{(x+1)^2}\mu \geq 0. \end{align}\] Next, by 68 71 , \[P_2=P_1+\nabla_{\psi}u_n\nabla_{\psi}=P_1+\frac{\partial_zu_n}{u_{n-1}^2}\partial_z.\] Then for \(z\) large enough, it holds\[\begin{align} P_2e^{-\frac{z^2}{x+1}\mu}=&P_1e^{-\frac{z^2}{x+1}\mu} +\frac{\partial_zu_n}{u_{n-1}^2}\partial_ze^{-\frac{z^2}{x+1}\mu}\\ \geq& e^{-\frac{z^2}{x+1}\mu} (\frac{1}{2} \frac{z^2}{(x+1)^2}\mu-C\frac{z}{x+1}\mu) \\ \geq &\frac{1}{4}e^{-\frac{z^2}{x+1}\mu} \frac{z^2}{(x+1)^2}\mu > 0. \end{align}\] ◻

Proof. By 223 and 58 , for \(\mu\) independent of \(\varepsilon\) and small enough such that \(\|\mu u_n\|_{L^\infty}\leq \frac{1}{8}\), when \(z\) is large, we have \[\begin{align} P_2 e^{-\frac{\psi_{n-1}^2(x,y,z)}{x+1}\mu}=&\nabla_\xi e^{-\frac{\psi_{n-1}^2(x,y,z)}{x+1}\mu} +(1+\tilde{q})\nabla_{\eta} e^{-\frac{\psi_{n-1}^2(x,y,z)}{x+1}\mu} -u_{n}\nabla_{\psi}^2e^{-\frac{\psi_{n-1}^2(x,y,z)}{x+1}\mu} \\ =&e^{-\frac{\psi_{n-1}^2(x,y,z)}{x+1}\mu} [ -(\tilde{q}\int_0^z \partial_yu_{n-1}dz'-\int_0^z \partial_yq_{n-1}dz')\frac{2\psi_{n-1}(x,y,z)}{x+1}\mu ]\\ &+e^{-\frac{\psi_{n-1}^2(x,y,z)}{x+1}\mu} [\frac{\psi_{n-1}^2(x,y,z)}{(x+1)^2}\mu -u_{n}(\frac{4\psi_{n-1}^2(x,y,z)}{(x+1)^2}\mu^2-\frac{2}{x+1}\mu) ]\\ \geq& 0. \end{align}\] ◻

Proof. By 70 and 223 , for \(\mu\) small enough and \(z\) big enough such that \(z\mu\gg1\), it holds \[\begin{align} P_1e^{-\frac{3}{2}\mu z^2}=& -\tilde{b}\partial_{z} e^{-\frac{3}{2}\mu z^2} -\frac{u_n}{u_{n-1}^2}\partial_{z}^2e^{-\frac{3}{2}\mu z^2} \\&+(-\frac{\partial_zu_{n}}{u_{n-1}^2} +\frac{u_{n}\partial_zu_{n-1}}{u_{n-1}^3}) (-3\mu z)e^{-\frac{3}{2}\mu z^2} \\ \leq &e^{-\frac{3}{2}\mu z^2} [ Cz\mu -\frac{u_n}{u_{n-1}^2}(9\mu^2 z^2-3\mu) ]\\ \leq& -e^{-\frac{3}{2}\mu z^2}< 0. \end{align}\] ◻

Proof. First, we consider the estimates near \(z=0.\) By \(|u_{n}-\bar{u}|\leq \varepsilon e^{-\frac{A}{x+1}}C{\epsilon_0}^2,\,\,|\partial_z (u_{n}-\bar{u})|\leq \varepsilon e^{-\frac{A}{x+1}}C{\epsilon_0}\) at \(z=0\) and \(|\partial_z^2 u_{n}-\partial_z^2 \bar{u}|\leq \varepsilon e^{-\frac{A}{x+1}}\) in \(\Omega\cap\{0\leq y\leq Y^*\}\), we have \[\begin{align} \label{26-06-13-1} |\partial_z (u_{n}-\bar{u})|\leq \varepsilon e^{-\frac{A}{x+1}}C(z+{\epsilon_0}), \quad | u_{n}-\bar{u}|\leq \varepsilon e^{-\frac{A}{x+1}} C(z+{\epsilon_0})^2\quad \text{in}\quad \Omega\cap\{0\leq y\leq Y^*\}. \end{align}\tag{231}\] Then we have, in \(\Omega\cap\{0\leq y\leq Y^*\},\) \[\begin{align} \begin{aligned} |\nabla_{n}\bar{u}^2|=|2\partial_z\bar{u}|\leq C, \quad |\partial_z\nabla_{n}\bar{u}^2|=|2\partial_z^2\bar{u}|\leq &C,\\ |\frac{ \partial_z(\bar{u}-u_{n-1})}{u_{n-1}}- \frac{ \partial_zu_{n-1}(\bar{u}-u_{n-1})}{u_{n-1}^2}|\leq& \varepsilon e^{-\frac{A}{x+1}}C , \end{aligned} \end{align}\] which gives ?? by 229 , the decay rates of derivatives of \(\bar{u}\) for large \(z\) and the definition of \(Y^*\).

By the definition of \(\nabla_\psi,\) we have \[\begin{align} \label{sm2}\begin{aligned} \nabla_\psi (u_{n}^2-\bar{u}^2)=\frac{1}{u_{n-1}}\partial_z (u_{n}^2-\bar{u}^2) =\frac{2}{u_{n-1}}(u_{n}\partial_z u_{n}-\bar{u}\partial_z \bar{u}). \end{aligned} \end{align}\tag{232}\] By 231 , we have \[\begin{align} \label{xinzeng1} |2(\frac{u_{n}}{u_{n-1}}\partial_z u_{n}-\frac{\bar{u}}{u_{n-1}}\partial_z \bar{u})|\leq \varepsilon e^{-\frac{A}{x+1}}C{\epsilon_0},\quad \text{at} \quad z=0. \end{align}\tag{233}\] Now we estimate \[\begin{align} \label{xinzeng}\begin{aligned} \partial_z\big( \frac{1}{u_{n-1}}(u_{n}\partial_z u_{n}-\bar{u}\partial_z \bar{u})\big)=&-\frac{\partial_zu_{n-1}}{u_{n-1}^2}(u_{n}\partial_z u_{n}-\bar{u}\partial_z \bar{u})\\&+\frac{1}{u_{n-1}}((\partial_z u_{n})^2-(\partial_z \bar{u})^2+u_{n}\partial_z^2 u_{n}-\bar{u}\partial_z^2\bar{u}).\end{aligned} \end{align}\tag{234}\] Again, since \(|u_{n}-\bar{u}|\leq \varepsilon e^{-\frac{A}{x+1}}C{\epsilon_0}^2, |\partial_z (u_{n}-\bar{u})|\leq \varepsilon e^{-\frac{A}{x+1}}C{\epsilon_0}\) at \(z=0\) and \(|\partial_z^2 u_{n}-\partial_z^2 \bar{u}|\leq \varepsilon e^{-\frac{A}{x+1}}\) in \(\Omega\cap\{0\leq y\leq Y^*\}\), we have \[\begin{align} | u_{n}\partial_z u_{n}-\bar{u}\partial_z \bar{u}|=& |u_{n}(\partial_z u_{n}-\partial_z \bar{u})+(u_{n}-\bar{u})\partial_z \bar{u}| \leq \varepsilon e^{-\frac{A}{x+1}}C(z+{\epsilon_0})^2 , \end{align}\] and \[\begin{align} |(\partial_z u_{n})^2-(\partial_z \bar{u})^2+u_{n}\partial_z^2 u_{n}-\bar{u}\partial_z^2\bar{u}|\leq \varepsilon e^{-\frac{A}{x+1}}C(z+{\epsilon_0}). \end{align}\] Hence, by 234 , we have \(|\partial_z \big(\frac{1}{u_{n-1}}(u_{n}\partial_z u_{n}-\bar{u}\partial_z \bar{u})\big)|\leq \varepsilon e^{-\frac{A}{x+1}}C.\) Combining this with 233 , we have \[\begin{align} |\frac{u_{n}}{u_{n-1}}\partial_z u_{n}-\frac{\bar{u}}{u_{n-1}}\partial_z \bar{u}|\leq \varepsilon e^{-\frac{A}{x+1}}C(z+{\epsilon_0}),\quad \text{in}\quad \Omega\cap\{0\leq y\leq Y^*\}. \end{align}\] Substituting this back to 232 , we obtain ?? . ◻

Proof. Set \(G=(z+c)f-\int_0^zf dz'.\) Then \(G(0)=cf(0)\), \(\partial_z G=(z+c)\partial_zf\) and \[|\partial_z G|\leq 2\varepsilon (z+c)^{1+\alpha}, \quad z>0,\] which implies \[|G|\leq\frac{\varepsilon (z+c)^{2+\alpha}}{1+\frac{\alpha}{2}}+c|f(0)| \quad \text{ for}\quad z>0.\] Hence, \[|\frac{G}{z+c}|\leq\frac{\varepsilon (z+c)^{1+\alpha}}{1+\frac{\alpha}{2}}+\frac{c}{z+c}|f(0)|\quad \text{ for}\quad z>0.\] ◻

Proof. For any fixed \((x,y)\in[0,X]\times[0,Y^*]\), take \[\begin{align} f=\partial_{z}(u_{n-1}-u_n)(x,y,z), \quad c=\frac{u_n(x,y,0)}{\partial_zu_n(x,y,0)} \end{align}\] in Lemma 19. By 235 , \(|\partial_z f|\leq |\partial_z^2(u_n-\bar{u})|+|\partial_z^2(u_{n-1}-\bar{u})|\leq2 \varepsilon e^{-\frac{A}{x+1}} (z+c)^\alpha\). Hence, by Lemma 19, we have \[\begin{align} \begin{aligned} &\frac{1}{z+\frac{u_n(x,y,0)}{\partial_zu_n(x,y,0)}}[\partial_{z}(u_{n-1}-u_n) -\frac{(u_{n-1}-u_n)-(u_{n-1}-u_n)|_{z=0}}{z+\frac{u_n(x,y,0)}{\partial_zu_n(x,y,0)}}]\\ \leq & \frac{\varepsilon e^{-\frac{A}{x+1}}(z+c)^{\alpha}}{1+\frac{\alpha}{2}}+\frac{1}{z+c}|f(0)| \\ \leq & \frac{\varepsilon e^{-\frac{A}{x+1}}(z+c)^{\alpha}}{1+\frac{\alpha}{3}} \quad \text{in}\quad \Omega\cap\{0\leq y\leq Y^*\},\end{aligned} \end{align}\]where we have used \(|f(0)|\leq \varepsilon^7 e^{-\frac{A}{x+1}}C\epsilon_0^2\) by the compatible approximate boundary data. ◻

Proof. Take \[\tilde{c}=\frac{u_n(x,y,0)}{\partial_zu_n(x,y,0)}.\] First, we investigate \((\frac{u_n}{u_{n-1}}\partial_{z}u_n)|_{z=0}.\) Since \(\partial_z\bar{u}\geq c_0\) at \(z=0\), \(|\partial_z^2(u_{n}-\bar{u})|\leq 1\) in \([0,Y^*]\) by the bootstrap assumption and \(|\partial_z^2(u_{n-1}-\bar{u})|\leq 1\) by the induction assumption, we have, by the boundary data at \(z=0\), that for some positive constants \(k_4\ll\alpha\) and \(\delta_5\) independent of \(\varepsilon,\) \[\begin{align} \label{dzunun}\begin{aligned} \partial_{z}u_n(1-k_4)\leq&(\frac{u_n}{u_{n-1}}\partial_{z}u_n)|_{z=0}\leq \partial_{z}u_n(1+k_4),\\ (1-k_4)\frac{1}{z+\tilde{c}}\leq&\frac{\partial_{z}u_n}{u_n}= \frac{\partial_{z}u_n}{\int_0^z \partial_{z}u_ndz'+u_n|_{z=0}}\leq (1+k_4)\frac{1}{z+\tilde{c}},\end{aligned} \end{align}\tag{236}\] in \(\Omega\cap\{y\leq Y^*\}\cap \{z\leq \delta_5\}\) which imply \[\begin{align} \label{1117-3} (1-k_4)^2\frac{1}{z+\tilde{c}}\leq (\frac{u_n}{u_{n-1}}\partial_{z}u_n)|_{z=0}\frac{1}{u_n}\leq (1+k_4)^2\frac{1}{z+\tilde{c}}, \end{align}\tag{237}\] in \(\Omega\cap\{y\leq Y^*\}\cap \{z\leq \delta_5\}.\) In fact, the second line in 236 comes from the following calculation. For any fixed \((x,y)\in[0,X]\times[0,Y^*],\) set \(b=u_n(x,y,0), a=\partial_zu_n(x,y,0)\geq c_0\) by 47 . Note \(\tilde{c}=\frac{b}{a}\). For small \(z\), we have \[\begin{align} \label{1117-5}\begin{aligned} \frac{\partial_{z}u_n}{u_n}=& \frac{\partial_{z}u_n}{\int_0^z \partial_{z}u_ndz'+u_n|_{z=0}} = \frac{a(1+o(1))}{\int_0^z a(1+o(1))dz'+b}\\ =&(1+o(1))\frac{1}{\int_0^z (1+o(1))dz'+\tilde{c}} =(1+o(1))\frac{1}{z (1+o(1))+\tilde{c}}\\ =&(1+o(1))^2\frac{1}{z +\tilde{c}} =(1+o(1))\frac{1}{z +\tilde{c}}.\end{aligned} \end{align}\tag{238}\] Second, we note that \[\begin{align} \partial_{z}(\frac{u_{n-1}}{u_n} )=&\partial_{z}(\frac{u_{n-1}-u_n}{u_n} ) =\frac{\partial_{z}(u_{n-1}-u_n)}{u_n} -\frac{\partial_{z}u_n(u_{n-1}-u_n)}{u_n^2} \\ =&\frac{1}{u_n}\big(\partial_{z}(u_{n-1}-u_n) -\frac{\partial_{z}u_n(u_{n-1}-u_n)}{u_n}\big). \end{align}\] Then\[\begin{align} \label{666-1-0205}\begin{aligned} |\partial_{z}(\frac{u_{n-1}}{u_n} )|\leq&\frac{1}{u_n}|\partial_{z}(u_{n-1}-u_n) -\frac{\partial_{z}u_n\big((u_{n-1}-u_n)-(u_{n-1}-u_n)|_{z=0}\big)}{u_n}| \\&+\varepsilon^6 e^{-\frac{A}{x+1}}C\epsilon_0,\end{aligned} \end{align}\tag{239}\] by the compatible approximate boundary condition 202 at \(z=0.\) Set \(f=\partial_{z}(u_{n-1}-u_n).\) Then \[\begin{align} \label{1119-2}\begin{aligned} &\partial_{z}(u_{n-1}-u_n) -\frac{\partial_{z}u_n\big((u_{n-1}-u_n)-(u_{n-1}-u_n)|_{z=0}\big)}{u_n} \\ =&f- (1+o(1))\frac{1}{z+\tilde{c}}\int_0^z fdz'\\ =&(1+o(1))(f- \frac{1}{z+\tilde{c}}\int_0^z fdz')+o(1)f,\end{aligned} \end{align}\tag{240}\] where we have used 236 . Then by Corollary 2, we have, for some positive constant \(\delta_5\) independent of \(\varepsilon,\) \[\begin{align} \label{649}\begin{aligned} &\frac{\partial_z u_n}{u_n}| \partial_{z}(u_{n-1}-u_n)-\frac{\partial_{z}u_n}{u_n}\big((u_{n-1}-u_n)-(u_{n-1}-u_n)|_{z=0}\big)| \\ \leq& \frac{\varepsilon e^{-\frac{A}{x+1}}(z+\tilde{c})^{\alpha}}{1+\frac{\alpha}{4}} \quad \text{in}\quad \Omega\cap\{0\leq y\leq Y^*\}\cap\{z\leq \delta_5\}.\end{aligned} \end{align}\tag{241}\] For the estimate on \(o(1)f,\) by the boundary condition 202 , 235 and integrating with respect to \(z\), we have\[\begin{align} |f|=|\partial_{z}(u_n-u_{n-1})|\leq \varepsilon^7 e^{-\frac{A}{x+1}}C\epsilon_0^2+\frac{2}{1+\alpha}\varepsilon e^{-\frac{A}{x+1}} (z+\tilde{c})^{1+\alpha}. \end{align}\] Combining 236 , 239 and 241 , we complete the proof of the lemma. ◻

Proof. Set \(f=\partial_x\bar{u}-\partial_xu_{n-1}\) and \(\tilde{c}=\frac{u_{n-1}(x,y,0)}{\partial_zu_{n-1}(x,y,0)}.\) By the compatible boundary data 202 and \(\partial_zu_B>0\) at \(z=0,\) we have \[\begin{align} \label{1119-3}\begin{aligned} \tilde{c} =&\frac{ u_{n-1}}{\partial_zu_{n-1}}|_{z=0}= (1+R_{\epsilon_0}) \frac{\bar{u}}{\partial_zu_{n-1}}|_{z=0}\\ =& (1+R_{\epsilon_0})^2 \frac{\bar{u}}{\partial_z\bar{u}}|_{z=0} = (1+R_{\epsilon_0})^2 \frac{u_B}{\partial_zu_B}|_{z=\epsilon_0} \\ =& (1+R_{\epsilon_0})^3 \epsilon_0,\end{aligned} \end{align}\tag{242}\] where \(R_{\epsilon_0}\) stands for the terms satisfying \(|R_{\epsilon_0}|\leq C\epsilon_0\) which may vary from line to line. Then by the induction assumption and 242 , we have, for \(f=\partial_{x}u_{n-1}-\partial_{x}\bar{u},\) it holds \[\begin{align} \label{0126huahua}\begin{aligned} |\partial_z f|=&|\partial_z\partial_{x}u_{n-1}-\partial_z\partial_{x}\bar{u}| \leq \varepsilon^2e^{-\frac{A}{x+1}} (\frac{1}{x+1})^{\frac{\alpha}{2}}(z+{\epsilon_0})^{\alpha} \\ \leq& \varepsilon^2e^{-\frac{A}{x+1}} (\frac{1}{x+1})^{\frac{\alpha}{2}}(z+\tilde{c}(1+C\epsilon_0))^{\alpha} ,\quad \Omega\cap \{\frac{z+{\epsilon_0}}{\sqrt{x+1}}\leq \delta\}.\end{aligned} \end{align}\tag{243}\] Then similar to the estimate 240 , by Remark 20 and 242 , for some \(\delta_6\ll\delta\) independent of \(\varepsilon,\) we have, in \(\Omega\cap \{z\leq \delta_6\},\) \[\begin{align} &\frac{\partial_z \bar{u}}{u_{n-1}}| ( \partial_x\bar{u}-\partial_xu_{n-1})-\frac{\partial_zu_{n-1}}{u_{n-1}} \int_0^z\partial_x\bar{u}-\partial_xu_{n-1}dz' |\\ =&|(1+o(1))\frac{1}{z+\tilde{c}}(f- (1+o(1))\frac{1}{z+\tilde{c}}\int_0^z fdz')|\\ =&|(1+o(1))^2\frac{1}{z+\tilde{c}}(f- \frac{1}{z+\tilde{c}}\int_0^z fdz')+o(1)\frac{f}{z+\tilde{c}}| \\ \leq &\frac{ \varepsilon^2}{2+\frac{\alpha}{2}} (\frac{1}{x+1})^{\frac{\alpha}{2}}e^{-\frac{A}{x+1}} (z+\epsilon_0)^{\alpha}, \end{align}\] where we have used \(\frac{\partial_z u_{n-1}}{u_{n-1}}=(1+o(1))\frac{1}{z+\tilde{c}}\) by 238 , \[\begin{align} \frac{\partial_z \bar{u}}{u_{n-1}}=&\frac{(1+o(1))\partial_z u_{n-1}}{u_{n-1}} =(1+o(1))\frac{1}{z+\tilde{c}}, \end{align}\]and 243 to estimate \(|o(1)\frac{f}{z+\tilde{c}}|.\) Hence, in \(\Omega\cap \{z\leq \delta_6\},\) \[\begin{align} |\partial_z \bar{u}\partial_z(\frac{\int_0^z\partial_x\bar{u}-\partial_xu_{n-1}dz'}{u_{n-1}})| =&\frac{\partial_z \bar{u}}{u_{n-1}}| ( \partial_x\bar{u}-\partial_xu_{n-1})-\frac{\partial_zu_{n-1}}{u_{n-1}} \int_0^z\partial_x\bar{u}-\partial_xu_{n-1}dz' |\\ \leq & \frac{ \varepsilon^2}{2+\frac{\alpha}{2}} (\frac{1}{x+1})^{\frac{\alpha}{2}}e^{-\frac{A}{x+1}} (z+\epsilon_0)^{\alpha}. \end{align}\] ◻

Proof. By 227 , \[\begin{align} \begin{aligned} -\partial_z\nabla_{\tau_1}\bar{u}+\partial_z\nabla_{\xi}\bar{u}=&(\frac{\int_0^z\partial_x\bar{u}dz'}{\bar{u}}-\frac{\int_0^z\partial_x u_{n-1}dz'}{u_{n-1}})\partial_z^2\bar{u} \\ &+\partial_z(\frac{(u_{n-1}-\bar{u})\int_0^z\partial_x\bar{u}dz'}{u_{n-1}\bar{u}})\partial_z \bar{u}+\partial_z(\frac{\int_0^z\partial_x\bar{u}-\partial_xu_{n-1}dz'}{u_{n-1}})\partial_z \bar{u}.\end{aligned} \end{align}\] By 205 , we have \[\begin{align} & |\partial_z^2\bar{u}|\leq C(z+\epsilon_0),\quad |\frac{\int_0^z\partial_x\bar{u}dz'}{u_{n-1}\bar{u}}|\leq C,\quad |\partial_z\frac{\int_0^z\partial_x\bar{u}dz'}{u_{n-1}\bar{u}}|\leq \frac{C}{\bar{u}}, \\&|u_{n-1}-\bar{u}|\leq\varepsilon^6 \phi_{1,1+2\alpha},\quad |\partial_zu_{n-1}-\partial_z\bar{u}|\leq \varepsilon^5 \phi_{1,\alpha}. \end{align}\] Then by Lemma 21, it holds that \[\begin{align} \begin{aligned} | \partial_z\nabla_{\tau_1}\bar{u}-\partial_z\nabla_{\xi}\bar{u}| \leq \varepsilon^3 \phi_{1,\alpha}C+\frac{ \varepsilon^2}{2+\frac{\alpha}{2}} (\frac{1}{x+1})^{\frac{\alpha}{2}}e^{-\frac{A}{x+1}} (z+\epsilon_0)^{\alpha},\quad \Omega\cap \{\sqrt{z+\epsilon_0}\leq \varepsilon^3 \},\end{aligned} \end{align}\] where we have used for the small positive constant \(\alpha,\) \[\begin{align} |\partial_z^2\bar{u}|\leq C(z+\epsilon_0)^{1-\alpha}(z+\epsilon_0)^\alpha \leq C\varepsilon^3(z+\epsilon_0)^\alpha\quad \text{in}\quad \Omega\cap\{z+\epsilon_0\leq \varepsilon^6 \}. \end{align}\] Note that for \(\varepsilon\) small, \(\sqrt{z+\epsilon_0}\leq \varepsilon^3\) implies \(z\leq \delta_6\) where \(\delta_6\) is defined in Lemma 21 which is independent of \(\varepsilon\). Similarly, we can derive \[\begin{align} \begin{aligned} | \partial_z\nabla_{\tau_1}\bar{u}-\partial_z\nabla_{\eta}\bar{u}| \leq \varepsilon^3 \phi_{1,\alpha}C+\frac{ \varepsilon^2}{2+\frac{\alpha}{2}} (\frac{1}{x+1})^{\frac{\alpha}{2}}e^{-\frac{A}{x+1}} (z+\epsilon_0)^{\alpha},\quad \Omega\cap \{\sqrt{z+\epsilon_0}\leq \varepsilon^3 \}.\end{aligned} \end{align}\] ◻